problem with a mysql query with data from a form

----- Original Message -----
From: "Vincent Fievet" <vincent.fievet@technocite.be>
To: <php-general@lists.php.net>
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 : 'php-general@lists.php.net'
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";
++++++++++++++++++
should be

$sql="SELECT * FROM journal WHERE a like '$det %' OR de like '$det %' order
by dateheure desc";
No need of prefixing $det with %
(blank space after $det.)
++++++++++++++++
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

-----Message d'origine-----
De : Vincent Fievet
Envoyé : mercredi 27 août 2003 10:44
À : 'php-general@lists.php.net'
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

You might also want to use addslashes() .

like '%something%' puts a big load on the database server. and it seems
you are doing mulitples of these queries. Please make sure you have the
correct indices on your tables. You might want to look at fulltext
indexing instead.

all the best


murugesan wrote:

>use
>$det=trim($det);
>then assign the $sql value.
>-murugesan
>
>----- Original Message -----
>From: "Vincent Fievet" <vincent.fievet@technocite.be>
>To: <php-general@lists.php.net>
>Sent: Friday, August 29, 2003 11:55 AM
>Subject: [php] problem with a mysql query with data from a form
>
>
>
>-----Message d'origine-----
>De : Vincent Fievet
>Envoyé : mercredi 27 août 2003 10:44
>À : 'php-general@lists.php.net'
>Objet : problem with a mysql query with data from a form
>
>
>hi,
>
>i am a newbie to php and mysql, i run the easyphp kit ( php 4.2.0, Mysql
>3.23.49 and phpmyadmin 2.2.6 )
>
>i would like to use a select * from mytable where my variable like
>'%$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
>
>
>


--
http://www.raditha.com/php/progress.php
A progress bar for PHP file uploads.