[PHP] One more time (Duft Punk) about hyperlink
Thank you very much, I do appreciate your help.
Ur script doens work, but, curiously, it works when I changed a bit...
select left(loc1,3) as locx, openart_id, tit1, pub1, id1, ref1 from
openart_table
If possible, can I ask one more thing?
This part is the problem
A HREF="/momatlib/image/? echo $row["locx"]; ?.jpg"? echo
$row["locx"]; ?/A
I got 3 letters, but I would like to show all "loc1" data (eg "1-2 near
entrance") in the table, and have first 3 letters of "loc1" for jpg
hyperlink (http://www.heyjude.com/image/1-2.jpg).
Do you know how to fix this?
script--
$result = mysql_query("select left(loc1,3) as locx, openart_id, tit1, pub1,
id1, ref1 from openart_table
where tit1 like '%$tit%'
and res1 like '%$res%'
and pub1 like '%$pub%'
and date1 like '%$date%'
and id1 like '%$id%'
and ref1 like '%$ref%'
and loc1 like '%$loc%'
and type1 like '%$type%'
and vol1 like '%$vol%'
and fre1 like '%$fre%'
and note1 like '%$note%' order by tit1");
$rows = mysql_num_rows($result);
echo $rows,"Record Found p";
while($row = mysql_fetch_array($result)){
?
tr
tda href = "openart_detail.php ?iden=? echo $row["openart_id"] ?"?
echo $row["openart_id"]; ?/abr/td
td? echo $row["tit1"]; ?br/td
td? echo $row["pub1"]; ?br/td
td? echo $row["id1"]; ?br/td
td? echo $row["ref1"]; ?br/td
td
A HREF="/momatlib/image/? echo $row["locx"]; ?.jpg"? echo
$row["locx"]; ?
/Abr
/td
/tr
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[PHP] PHP script problem (with MySQL)
Hello,
I need your help about PHP (ver 4.3.1) and MySQL (ver 3.23) database.
I am struggling to customise a database (using left command etc), and am
stuck now.
Please have a brief look at my scripts below.
What I do like to do is to get first 3 letters from "loc1" column from
openart_table (MySQL).
And show the data in the follwoing part...
td? echo $row["loc1"]; ?br/td
but also I would like to make hyperlinks to jpg files to images/"the first 3
letters from loc1".jpg
So it will be something like this...
td
a href="/image/? echo $row["loc1"]; ?.jpg"? echo $row["loc1"];
?/abr
/td
But when I tried, "select left (loc1,3) from openart_table where.."
doesnt work properly.
I do not want to destroy anything for the rest of the table.
I need all data from "tit1", "pub1", "id1", and "ref1" in each column.
Could you show me a script example?
Could you help me, please? Thank you.
-PHP script from here
html
body
table border="1" bordercolor="black" cellspacing="0" align="center"
BGCOLOR="#CC" width="900"
tr bgcolor="#ffcc66"
td align="center"bID No/bbr/td
td align="center" BGCOLOR="#CC3366"bBook Title/bbr/td
td align="center"bPublisher/bbr/td
td align="center"bID/bbr/td
td align="center"bReference/bbr/td
td align="center" width="10%"bLocation/bbr/td
/tr
?
mysql_connect(localhost,root,cheers);
mysql_select_db(openart);
if($tit == ""
$res == ""
$pub == ""
$date == ""
$id == ""
$ref == ""
$loc == ""
$type == ""
$vol == ""
$fre == ""
$note == "")
{
echo 'Type something';
}
elseif($tit == "%" || $res == "%" || $pub == "%" || $date == "%" || $id ==
"%" ||
$ref == "%" || $loc == "%" || $type == "%" || $vol == "%" || $fre ==
"%" || $note == "%"){
echo 'Not Valid';
}
else{
if($tit == ""){
$tit = '%';
}
if($res == ""){
$res = '%';
}
if($pub == ""){
$pub = '%';
}
if($date == ""){
$date = '%';
}
if($id == ""){
$id = '%';
}
if($ref == ""){
$ref = '%';
}
if($loc == ""){
$loc = '%';
}
if($type == ""){
$type = '%';
}
if($vol == ""){
$vol = '%';
}
if($fre == ""){
$fre = '%';
}
if($note == ""){
$note = '%';
}
$result = mysql_query("select * from openart_table
where tit1 like '%$tit%'
and res1 like '%$res%'
and pub1 like '%$pub%'
and date1 like '%$date%'
and id1 like '%$id%'
and ref1 like '%$ref%'
and loc1 like '%$loc%'
and type1 like '%$type%'
and vol1 like '%$vol%'
and fre1 like '%$fre%'
and note1 like '%$note%' order by tit1");
$rows = mysql_num_rows($result);
echo $rows,"Records Found p";
while($row = mysql_fetch_array($result)){
?
tr
tda href = "openart_detail.php ?iden=? echo $row["openart_id"] ?"?
echo $row["openart_id"]; ?/abr/td
td? echo $row["tit1"]; ?br/td
td? echo $row["pub1"]; ?br/td
td? echo $row["id1"]; ?br/td
td? echo $row["ref1"]; ?br/td
td? echo $row["loc1"]; ?br/td
/tr
?
}
}
?
/table
/body
/html
end of script
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[PHP] Thanks but problem again! Re: [PHP] PHP script problem (with MySQL)
Thank you, Richard, I simply change this part (select * from openart_table) into (select substring(loc1, 1, 3) as THREE_LETTER_CODE, * from openart_table) but Ive got this error, whats wrong with it? Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in c:\apache..\openart\test\openart_search.php on line 98 Records Found Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\apache\openart\test\openart_search.php on line 100 Cheers -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
