Re: [PHP-DB] Prepared Statement Insert Problem
Remove the quotes around the variables in all your statements.
For example, this statement:
mysqli_stmt_bind_param($submitadmin, isss, '$numrows', '$admin',
sha1('$password'), '$email');
could be rewritten as:
mysqli_stmt_bind_param($submitadmin, isss, $numrows, $admin,
sha1($password), $email);
On Tue, Jul 21, 2009 at 8:01 PM, Jason Carson ja...@jasoncarson.ca wrote:
Hello everyone, I have a problem.
I use the following to *try* and insert data into my MySQL database...
//Variables come from a form
$username= $_POST['username'];
$password = $_POST['password'];
$email = $_POST['email'];
//Connect to the database
$connect = mysqli_connect($hostname, $dbusername, $dbpassword,
$database)or die(cannot connect);
//Find out how many rows in the database
$aidcount = mysqli_query ($connect, SELECT * FROM administrator);
$numrows = mysqli_num_rows($aidcount);
//The next 3 lines are using prepared statements to insert data but the
//second line ...mysqli_stmt_bind_param.. results in this error...
//Fatal error: Only variables can be passed by reference in file.php line
46
$submitadmin = mysqli_prepare($connect2, INSERT INTO administrator VALUES
(?, ?, ?, ?));
mysqli_stmt_bind_param($submitadmin, isss, '$numrows', '$admin',
sha1('$password'), '$email');
mysqli_stmt_execute($submitadmin);
...anyone know how I can solve this problem so I can insert data into my
database with prepared statements?
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Re: [PHP-DB] Prepared Statement Insert Problem
That worked, thanks!
Remove the quotes around the variables in all your statements.
For example, this statement:
mysqli_stmt_bind_param($submitadmin, isss, '$numrows', '$admin',
sha1('$password'), '$email');
could be rewritten as:
mysqli_stmt_bind_param($submitadmin, isss, $numrows, $admin,
sha1($password), $email);
On Tue, Jul 21, 2009 at 8:01 PM, Jason Carson ja...@jasoncarson.ca
wrote:
Hello everyone, I have a problem.
I use the following to *try* and insert data into my MySQL database...
//Variables come from a form
$username= $_POST['username'];
$password = $_POST['password'];
$email = $_POST['email'];
//Connect to the database
$connect = mysqli_connect($hostname, $dbusername, $dbpassword,
$database)or die(cannot connect);
//Find out how many rows in the database
$aidcount = mysqli_query ($connect, SELECT * FROM administrator);
$numrows = mysqli_num_rows($aidcount);
//The next 3 lines are using prepared statements to insert data but the
//second line ...mysqli_stmt_bind_param.. results in this error...
//Fatal error: Only variables can be passed by reference in file.php
line
46
$submitadmin = mysqli_prepare($connect2, INSERT INTO administrator
VALUES
(?, ?, ?, ?));
mysqli_stmt_bind_param($submitadmin, isss, '$numrows', '$admin',
sha1('$password'), '$email');
mysqli_stmt_execute($submitadmin);
...anyone know how I can solve this problem so I can insert data into my
database with prepared statements?
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Re: [PHP-DB] Prepared Statement Insert Problem
kesavan trichy rengarajan wrote: could be rewritten as: mysqli_stmt_bind_param($submitadmin, isss, $numrows, $admin, sha1($password), $email); Turning on E_STRICT in PHP 5.3 will show PHP Strict Standards: Only variables should be passed by reference This is also true in earlier versions although the warning isn't displayed. For compliance, try: $s = sha1($password); mysqli_stmt_bind_param($submitadmin, isss, $numrows, $admin, $s, $email); Chris -- Blog: http://blogs.oracle.com/opal Twitter: http://twitter.com/ghrd -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Prepared Statement Insert Problem
Done. Thanks for letting me know about that. kesavan trichy rengarajan wrote: could be rewritten as: mysqli_stmt_bind_param($submitadmin, isss, $numrows, $admin, sha1($password), $email); Turning on E_STRICT in PHP 5.3 will show PHP Strict Standards: Only variables should be passed by reference This is also true in earlier versions although the warning isn't displayed. For compliance, try: $s = sha1($password); mysqli_stmt_bind_param($submitadmin, isss, $numrows, $admin, $s, $email); Chris -- Blog: http://blogs.oracle.com/opal Twitter: http://twitter.com/ghrd -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Prepared Statement Insert Problem
Hi i wanna ask a question. I am trying to create an image on fly, please do
help me , following is the code.
*File Name : Font.php
Code: *
html
head
titleImage Creation/title
script language=javascript
var xmlhttp;
function showPic()
{
xmlhttp=GetXmlHttpObject();
if (xmlhttp==null)
{
alert (Browser does not support HTTP Request);
return;
}
var text = document.getElementById(textfield).value;
var url=image.php;
url=url+?text=text;
alert(url);
url=url+sid=+Math.random();
xmlhttp.onreadystatechange=stateChanged;
xmlhttp.open(GET,url,true);
xmlhttp.send(null);
}
function stateChanged()
{
if (xmlhttp.readyState==4)
{
if(xmlhttp.responseText == 1)
{
document.getElementById(pic).style.display=block;
}
}
}
function GetXmlHttpObject()
{
if (window.XMLHttpRequest)
{
// code for IE7+, Firefox, Chrome, Opera, Safari
return new XMLHttpRequest();
}
if (window.ActiveXObject)
{
// code for IE6, IE5
return new ActiveXObject(Microsoft.XMLHTTP);
}
return null;
}
/script
/head
body
form name=form1 method=post action= onSubmit= return false;
table width=304 border=1
tr
td colspan=2div align=centerText/div/td
/tr
tr
tdText/td
tdlabel
input type=text name=textfield id=textfield
/label/td
/tr
tr
td colspan=2label
div align=center
input type=submit name=button id=button value=Update
onClick=showPic()
/div
/label/td
/tr
tr
td colspan=2div id=pic style=display:noneimg src=pic.jpg
//div/td
/tr
/table
/form
/body
/html
*File Name : image.php
Code:*
[php]
$name = $_GET['text'];
$pic = imagecreatetruecolor(100, 100);
$text_color = imagecolorallocate($pic, 255, 255, 255);
imagestring($pic, 10, 15, 15, $name, $text_color);
$pi = Imagejpeg($pic,pic.jpg);
echo $pi;
ImageDestroy($pic);
[/php]
*Problem: *
What this code is doing is that, it creates a new image with the text (that
user enters) on it, but loads the image that was created previously, I want
that it should display the text on the picture which users enter on the fly.
( e.g. If user enter TEXT as text it should display TEXT on the picture
displayed).Hope you got the problem.
Thanks
For me, it shoould work like this: User enters text intext filed (in
font.html) , when pressed button, image.php should create an im
On Tue, Jul 21, 2009 at 10:46 PM, Christopher Jones
christopher.jo...@oracle.com wrote:
kesavan trichy rengarajan wrote:
could be rewritten as:
mysqli_stmt_bind_param($submitadmin, isss, $numrows, $admin,
sha1($password), $email);
Turning on E_STRICT in PHP 5.3 will show
PHP Strict Standards: Only variables should be passed by reference
This is also true in earlier versions although the warning isn't
displayed.
For compliance, try:
$s = sha1($password);
mysqli_stmt_bind_param($submitadmin, isss, $numrows, $admin, $s,
$email);
Chris
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Blog: http://blogs.oracle.com/opal
Twitter: http://twitter.com/ghrd
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