use
$det=trim($det);
then assign the $sql value.
-murugesan
----- Original Message -----
From: "Vincent Fievet" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Friday, August 29, 2003 11:55 AM
Subject: [PHP] problem with a mysql query with data from a form
-----Message d'origine-----
De : Vincent Fievet
Envoyé : mercredi 27 août 2003 10:44
À : '[EMAIL PROTECTED]'
Objet : problem with a mysql query with data from a form
hi,
i am a newbie to php and mysql, i run the easyphp kit ( php 4.2.0, Mysql
3.23.49 and phpmyadmin 2.2.6 )
i would like to use a select * from mytable where my variable like
'%$mystring%'
where $mystring come from a form
<form action="show.php" method="post" name="form1" target="_blank">
<table align="left">
<tr> '%Mike %
<td width="98"><h3>Communication Impliquant :</h3></td>
<td width="144" valign="middle"> <font color="#FFFFCC">
<select name="det" size="5" id="det">
<option value="PC" selected>PC</option>
<option value="echo">echo</option>
...
<option value="H10 Tango">H10 Tango</option>
</select>
<select name="den" size="1" id="den">
<option value=" " selected> </option>
<option value="1">1</option>
...
<option value="10">10</option>
</select>
</font></td>
<td width="153"> <input name="delibre" type="text" id="delibre"
maxlength="20"></td>
<td width="39"> <input type="submit" name="Submit" value="Ok"> </td>
so ,after choosing "PC" in the form, in show.php i do a
$sql="SELECT * FROM journal WHERE a like '% $det%' OR de like '%$det%' order
by dateheure desc";
echo $sql;
and what i get is :
SELECT * FROM journal WHERE a like '%PC %' OR de like '%PC %' order by
dateheure desc
( note the blank space at the end of PC )
whitch doesnt get any record since the word i am looking is "PC" and not "PC
"
i used a "like" because the request should be used to find record with "PC
1" "PC" "PC 9" "PC12" ...
any ideas ?
Thanx
Vincent Fievet
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