The variable is being passed properly; it just doesn't have a value. Where are you setting $listbox??
---John Holmes... > -----Original Message----- > From: Chip Wiegand [mailto:[EMAIL PROTECTED]] > Sent: Tuesday, August 06, 2002 9:28 AM > To: 1LT John W. Holmes > Cc: php > Subject: Re: [PHP] passing select list variable from page1 to > insertstatement astable name on page2 > > On Mon, 2002-08-05 at 23:21, 1LT John W. Holmes wrote: > > Change this: > > > > mysql_query($sql) or die (mysql_error()); > > > > to this: > > > > mysql_query($sql) or die ("Error in this query >>$sql<< : " . > > mysql_error()); > > > > And the error will probably be obvious to you. $listbox is not getting a > > value. Why doesn't listbox appear in your second URL? > > > > ---John Holmes... > > Thanks for the better error statement. It does shed some more light on > the subject - the table-name variable is not being passed on submit of > the second page - > > Error in this query >>insert into (today,exercise,reps,comments) values > ('08-06','120','20','test')<< : You have an error in your SQL syntax > near '(today,exercise,reps,comments) values ('08-06','120','20','test')' > at line 1 > > No matter if I quote the variable or not, always the same error. The sql > code is just below a couple paragraphs, I am at a loss as to how to get > that variable to be passed properly. > > -- > Chip W > www.wiegand.org > [EMAIL PROTECTED] > > > ----- Original Message ----- > > From: "Chip Wiegand" <[EMAIL PROTECTED]> > > To: "php" <[EMAIL PROTECTED]> > > Sent: Tuesday, August 06, 2002 1:06 AM > > Subject: [PHP] passing select list variable from page1 to insert > statement > > astable name on page2 > > > > > > > > > > I have a page with a select list that will contain as many as 20 > > > options, allowing only one choice, not multiple. I have a second page > > > that is loaded on submit that wants that select list variable for the > > > table name in an insert statement. On the second page is a form for > > > entering data into the table, chosen from the select list on the first > > > page. > > > > > > I can echo the variable on the second page, so I know it is being sent > > > from page1 to page2 appropriately. I see in the url that it is being > > > send appropriately, and upon filling in the form, again the url shows > > > the data that is to be sent to the database. All appears to be okay, > > > except I get a syntax error message. > > > > > > Below is copied messages I have been exchanging with another user on > > > this subject, it contains a complete history, and the code for the two > > > pages, at the bottom. > > > > > > I hope someone can help solve this perplexing problem. > > > > > > -- > > > Chip W > > > www.wiegand.org > > > [EMAIL PROTECTED] > > > > > > Begin previous posts -- > > > > ======================================================================== = > > > > > > Well, I think we're getting closer, in that the correct data is being > > > sent, as seen here - > > > > > > http://192.168.1.53/test2.php?listbox=leglifts&exerciselist= > > > > > > Then when I enter some data and press submit the correct data is sent > - > > > > > > > > > http://192.168.1.53/test2.php?exercise=80&reps=12&comments=test&submit=S en > d+ > > Data > > > > > > But at the same time I get this error - > > > > > > You have an error in your SQL syntax near > > > '(date,exercise,reps,comments) values ('08-05','80','12','test')' at > > > line 1 > > > > > > My entire sql statement looks like this - > > > > > > <? > > > if(isset($submit)): > > > $db = mysql_connect("localhost","----","----"); > > > if(!$db) error_message(sql_error()); > > > mysql_select_db("workout",$db) or die (mysql_error()); > > > $date = date("m-d"); > > > $sql = "insert into $listbox (date,exercise,reps,comments) > > > values ('$date','$exercise','$reps','$comments')"; > > > mysql_query($sql) or die (mysql_error()); > > > endif; > > > echo $listbox; > > > ?> > > > > > > As for the variable $listbox in the insert statement, I have tried to > > > quote it - ".$listbox." - and also like this - "$listbox" - both of > > > which result in the same error as above, and then this version - > > > '$listbox' - causing a T-Variable error which crashes the page. > > > I don't see any syntax error in there. I will be posting this on the > > > lists also. > > > > > > -- > > > Chip W > > > www.wiegand.org > > > [EMAIL PROTECTED] > > > > > > On Mon, 2002-08-05 at 20:18, Rich Hutchins wrote: > > > > Chip, > > > > > > > > I'm going to take one more shot here, because, frankly, I'm stumped > > > too. > > > > > > > > If you look at your URL, you see: > > > > > > > > > http://192.168.1.53/workout- > absflexor.php?exerciselist=%24listbox&exercise=1 > > > > 20 > > > > &reps=12&date=1-1&comments=test&submit=Send+Data > > > > > > > > The part that reads ?exerciselist=%24listbox is what worries me. > > > > exerciselist should have the value from the listbox in it and > > > exerciselist > > > > is what you should be referencing in your sql. So this part of the > URL > > > > should look something like: > > > > > > > > ...?exerciselist=crunches&... > > > > > > > > Then you'd use a reference to $exerciselist in your SQL, not > $listbox. > > > > > > > > Forget what the book says for a moment. They're prone to errors just > > > like > > > > you and me. > > > > > > > > Variables are passed in name/value pairs. Everything to the left of > > > the = is > > > > a name and everything to the right is a value. Pretty simple. Your > URL > > > does > > > > not have a valid name/value pair for the variable $exerciselist. > > > > $exerciselist is valid, but %24listbox is not. Fix that and you're > > > home > > > > free. > > > > > > > > My hunch (can't you just hear Velma saying this to Fred?) is that > > > there's > > > > something wrong with the way you are attempting to assign the value > of > > > > listbox to exerciselist on your first page. I have never worked with > > > passing > > > > <select> values, so you're one up on me there. But if you look > > > closely, > > > > there has to be something funky going on there. > > > > > > > > If you can't figure it out, maybe post to this list or the php- > general > > > list > > > > a question about how to pass the values of <select>s from page to > > > page. You > > > > might also want to check this out too: > > > > > > > > http://www.php.net/manual/en/faq.html.php > > > > > > > > If you figure it out, please post the answer because I want to learn > > > too. > > > > > > > > Wait...A rock just hit me... > > > > > > > > Look at this: > > > > > > > > > > echo "<input type='hidden' name='exerciselist' > value='$listbox'>"; > > > > > > > > Know what you're saying here? The value of exerciselist is the > > > _STRING_ > > > > $listbox. Not the VALUE of the $listbox, the _LITERAL STRING_. Try > > > saying it > > > > this way: > > > > > > > > echo "<input type='hidden' name='exerciselist' > value='".$listbox."'>"; > > > > > > > > I don't know if this grabs the value of $listbox at submit time, but > > > it > > > > should escape you from the HTML being echoed and substitute the > value > > > of the > > > > variable $listbox. > > > > > > > > Now I'm really finished ;^) > > > > > > > > Good luck, > > > > Rich > > > > > > > > -----Original Message----- > > > > From: Chip Wiegand [mailto:[EMAIL PROTECTED]] > > > > Sent: Monday, August 05, 2002 12:53 AM > > > > To: Rich Hutchins > > > > Subject: RE: [PHP-DB] problem getting form input field to become > > > > partofquery statement > > > > > > > > > > > > On Sun, 2002-08-04 at 09:41, Rich Hutchins wrote: > > > > > Chip, > > > > > > > > > > Could you post the result of your echo statement please? > > > > > > > > Here's the error returned: > > > > > > > > You have an error in your SQL syntax near '(date, comments, reps, > > > > exercise) values('1-1','test','12','120')' at line 1 > > > > > > > > And this is the url that matches the above error: > > > > > > > > > > > > > http://192.168.1.53/workout- > absflexor.php?exerciselist=%24listbox&exercise=1 > > > > 20 > > > > &reps=12&date=1-1&comments=test&submit=Send+Data > > > > > > > > The error has to be the table name, it's wrong. The variable > $listbox > > > > isn't being used. > > > > I can echo $listbox (as seen below, last line) and it returns the > one > > > > value I want, and selected in the list box on the first page. > > > > Why the same variable doesn't work in the insert statement is what's > > > > bugging me. > > > > > > > > <? > > > > if(isset($submit)): > > > > $db = mysql_connect("localhost","----","-----"); > > > > if(!$db) error_message(sql_error()); > > > > mysql_select_db("workout",$db) or die (mysql_error()); > > > > //$date = date("m-d"); > > > > $sql = "insert into $listbox (date, comments, reps, exercise) > > > > values('$date','$comments','$reps','$exercise')"; > > > > mysql_query($sql) or die(mysql_error()); > > > > endif; > > > > echo $listbox; > > > > ?> > > > > > > > > > Judging from your code, it looks like your $exercises array > contains > > > three > > > > > table names in your db. When your user selects a value from the > > > listbox, > > > > you > > > > > assign that variable to the hidden field exerciselist. > > > > > > > > Eventually, the select list will contain aprox. 20 selections, the > > > user > > > > would pick one only, which is passed to the next page and becomes > the > > > > table name in the insert statement. If there's a better way to do > this > > > > than what I'm trying, I'm open to suggestions. > > > > > > > > > Then when the user hits submit, the value in exercise list gets > sent > > > to > > > > the > > > > > receiving page where the user should be able to enter additional > > > info > > > > > regarding that particular exercise. > > > > > > > > > > I'm curious then, why are you attempting to use the variable > > > $listbox as a > > > > > source for the table name in your SQL statement (I'm assuming > that's > > > what > > > > > $listbox is supposed to be doing for you) instead of the single > > > value that > > > > > gets passed via the exerciselist hidden field? > > > > > > > > I'm just following the examples is the book Beginning PHP4, by > > > > Choi/Kent/Lea. The show a select list with a name 'listbox' on the > > > first > > > > page, and the same name echoed on the second page. Simple enough. > > > > > > > > > Seems to me that $listbox would have the three values from the > array > > > > stored > > > > > in it but that $exerciselist would have only the one chosen from > the > > > > > listbox. I could see why your query would bomb since MySQL > wouldn't > > > know > > > > > which of the three tables you're trying to insert into. > > > > > > > > When I replace the variable $listbox with $exerciselist I get the > > > > following error: > > > > > > > > Table 'workout.$exerciselist' doesn't exist > > > > > > > > > Hope I'm not torturing you here. I certainly don't mean to. Just > > > trying to > > > > > help. > > > > > > > > No torturing, I need the help, thanks. > > > > > > > > > Rich > > > > > > > > -- > > > > Chip W > > > > www.wiegand.org > > > > [EMAIL PROTECTED] > > > > > > > > > > > > > > -----Original Message----- > > > > > From: Chip Wiegand [mailto:[EMAIL PROTECTED]] > > > > > Sent: Sunday, August 04, 2002 1:33 AM > > > > > To: Rich Hutchins > > > > > Cc: phpdb > > > > > Subject: RE: [PHP-DB] problem getting form input field to become > > > part > > > > > ofquery statement > > > > > > > > > > > > > > > On Sat, 2002-08-03 at 06:05, Rich Hutchins wrote: > > > > > > Try referencing the $listbox variable in you SQL statement like > > > this: > > > > > > > > > > > > $sql = "insert into ".$listbox." > > > > > > values(NULL,'$date','$exercise','$reps','$comments')"; > > > > > > > > > > > > I'm guessing that it might also work like this: > > > > > > > > > > > > $sql = "insert into '$listbox' > > > > > > values(NULL,'$date','$exercise','$reps','$comments')"; > > > > > > > > > > > > I think your core problem is that the $listbox variable is not > > > being > > > > > > evaluated properly in the SQL statement. Once you solve that, > > > you're > > > > good > > > > > to > > > > > > go. > > > > > > > > > > > > Hope this helps. > > > > > > > > > > > > Rich > > > > > > > > > > I tried both suggestions and neither are working. I am using the > get > > > > > format for the form so I can see what is being sent, and I am > > > getting > > > > > this: > > > > > > > > > > > > > > http://192.168.1.53/workout- > absflexor.php?exerciselist=%24listbox&exercise=8 > > > > > 0 > > > > > &reps=12&comments=&submit=Send+Data > > > > > > > > > > Notice that the $listbox variable is still not being sent. Just > > > above > > > > > this is an echo statement which shows what's in that variable, and > > > it > > > > > displays the expected result (see code below). I am at a loss as > to > > > why > > > > > this is not working. > > > > > > > > > > If I replace the $listbox variable with the table name shown in > the > > > echo > > > > > statement, a connection is made and the query is completed. > > > > > > > > > > -- > > > > > Chip W > > > > > www.wiegand.org > > > > > [EMAIL PROTECTED] > > > > > > > > > > > > > > > > -----Original Message----- > > > > > > From: Chip Wiegand [mailto:[EMAIL PROTECTED]] > > > > > > Sent: Saturday, August 03, 2002 12:51 AM > > > > > > To: phpdb > > > > > > Subject: [PHP-DB] problem getting form input field to become > part > > > of > > > > > > query statement > > > > > > > > > > > > > > > > > > I have a form with a select list and a hidden field to save the > > > selected > > > > > > item. On submit another page is loaded with a few fields to be > > > filled in > > > > > > and submitted to a table. Using get I see the data is being > passed > > > from > > > > > > the first page to the second properly, and the second page sends > > > its > > > > > > data properly. The hidden field from the first page is to be > used > > > by the > > > > > > second page as the name of the table in the query. I have a > place > > > where > > > > > > I echo the contents of the hidden field just to be sure it is > > > correct, > > > > > > and that does indeed show what I expect. I then make the query > > > statement > > > > > > point to the variable but it always responds that it cannot find > > > the > > > > > > table. The table does exist, the variable does contain the > > > appropriate > > > > > > table name, but is not being replaced by the name. What am I > doing > > > > > > wrong? > > > > > > > > > > > > -- > > > > > > Chip W > > > > > > www.wiegand.org > > > > > > [EMAIL PROTECTED] > > > > > > > > > > > > Below is the code for the first page -- > > > > > > > > > > > > <html> > > > > > > <head> > > > > > > <title></title> > > > > > > </head> > > > > > > <body> > > > > > > <div align="center"> > > > > > > <? > > > > > > $exercises=array("absflexor","absmachine","leglifts"); > > > > > > > > > > > > echo "<form action='workout-absflexor.php' method='get'>"; > > > > > > echo "<table width='70%' border='0' align='center'>"; > > > > > > echo "<tr><th align='center'><h2>Exercise Data > > > Input</h2></th></tr>"; > > > > > > echo "<tr><th align='center'><select name='listbox'>"; > > > > > > echo "<option>$exercises[0]</option>"; > > > > > > echo "<option>$exercises[1]</option>"; > > > > > > echo "<option>$exercises[2]</option>"; > > > > > > echo "</select><br><br>"; > > > > > > echo "</th></tr></table>"; > > > > > > echo "<input type='hidden' name='exerciselist' > value='$listbox'>"; > > > > > > echo "<input type='submit'>"; > > > > > > echo "</form>"; > > > > > > ?> > > > > > > </div> > > > > > > </body> > > > > > > </html> > > > > > > > > > > > > And below is the code for the second page -- > > > > > > > > > > > > <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" > > > > > > "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd";> > > > > > > <html xmlns="http://www.w3.org/1999/xhtml";> > > > > > > <head> > > > > > > <title>Workout Data Entry Form</title> > > > > > > <style type="text/css"> > > > > > > body { background-color: aqua; } > > > > > > div.c1 {text-align: center} > > > > > > </style> > > > > > > </head> > > > > > > <body> > > > > > > <div class="c1"> > > > > > > <h2>Work-Out Data Entry Screen</h2> > > > > > > <form action="<? PHP_SELF ?>" method="get"> > > > > > > <table summary="" width="60%" border="1" align="center" > > > > > > bgcolor="green"> > > > > > > <tr> > > > > > > <th>Weight</th> > > > > > > <td align="left"><input type="text" name="exercise" > > > > > > maxlength="4"></td> > > > > > > </tr> > > > > > > <tr> > > > > > > <th>Reps</th> > > > > > > <td align="left"><input type="text" name="reps" > > > > > > maxlength="4"></td> > > > > > > </tr> > > > > > > <tr> > > > > > > <th>Comments</th> > > > > > > <td colspan="2"><textarea cols="50" rows="3" > > > > > > name="comments"> > > > > > > </textarea></td> > > > > > > </tr> > > > > > > <tr><td><? echo $listbox; ?></td></tr> > > > > > > </table> > > > > > > <br /> > > > > > > <input type="submit" name="submit" value="Send Data" /> > > > <input > > > > > > type="reset" /> > > > > > > </form> > > > > > > </div> > > > > > > <? > > > > > > if(isset($submit)): > > > > > > $db = mysql_connect("localhost","----","------"); > > > > > > if(!$db) error_message(sql_error()); > > > > > > mysql_select_db("workout",$db) or die ("Ack! Where's the > > > database?"); > > > > > > $date = date("m-d"); > > > > > > $sql = "insert into $listbox > > > > > > values(NULL,'$date','$exercise','$reps','$comments')"; > > > > > > mysql_query($sql) or die ("Ack! No response when I queried the > > > > > > server!"); > > > > > > endif; > > > > > > ?> > > > > > > </body> > > > > > > </html> > > > > > > > > > > > > > > > > > > -- > > > PHP General Mailing List (http://www.php.net/) > > > To unsubscribe, visit: http://www.php.net/unsub.php > > > > > > > > > -- > > PHP General Mailing List (http://www.php.net/) > > To unsubscribe, visit: http://www.php.net/unsub.php > > > > > > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
