SQL ERROR

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sigridann
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SQL ERROR

Postby sigridann » Mon Apr 29, 2013 12:09 am

Hi all

i have this code that will search a record from the database

Code: Select all

<?php
define ('DB_NAME', 'try1');
define ('DB_USER', 'root');
define ('DB_PASSWORD', '');
define ('DB_HOST', 'localhost');
 
$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);

if (!$link) {
   die('Could not connect: ' . mysql_error());
}

$db_selected = mysql_select_db(DB_NAME, $link);

if (!$db_selected) {
   die('Can\'t use ' . DB_NAME . ': ' . mysql_error());
}

$term = $_GET['fname'];
 
$sql = mysql_query("select * from employee where fname like '%$term%'");
$result=mysql_query($sql);
$num=mysql_num_rows($result);

//mysql_close();

$i=0;
while ($i < $num) {
//while ($row = mysql_fetch_array($sql)){
    $emp_number=mysql_result($result,$i,"emp_number");
   $fname=mysql_result($result,$i,"fname");
   $mname=mysql_result($result,$i,"mname");
   $lname=mysql_result($result,$i,"lname");
   $homeadd=mysql_result($result,$i,"homeadd");
   $emailadd=mysql_result($result,$i,"emailadd");
   $ofcnum=mysql_result($result,$i,"ofcnum");
   $mobilenum=mysql_result($result,$i,"mobilenum");
   $homenum=mysql_result($result,$i,"homenum");
   $position=mysql_result($result,$i,"position");
   $practice=mysql_result($result,$i,"practice");
   $projname=mysql_result($result,$i,"projname");
   $projmgr=mysql_result($result,$i,"projmgr");
   $teamlead=mysql_result($result,$i,"teamlead");
    }
 
   ?>
<html><br />
<input name="textfield" type="text" value="<?php echo $emp_number; ?>" size="40" />
<br />
<input name="textfield" type="text" value="<?php echo $fname; ?>" size="40" />
<br />
<input name="textfield" type="text" value="<?php echo $mname; ?>" size="40" />
<br />
<input name="textfield" type="text" value="<?php echo $lname; ?>" size="40" />
<br />
<input name="textfield" type="text" value="<?php echo $homeadd; ?>" size="40" />
<br />
<input name="textfield" type="text" value="<?php echo $emailadd; ?>" size="40" />
<br />
<input name="textfield" type="text" value="<?php echo $ofcnum; ?>" size="40" />
<br />
<input name="textfield" type="text" value="<?php echo $mobilenum; ?>" size="40" />
<br />
<input name="textfield" type="text" value="<?php echo $homenum; ?>" size="40" />
<br />
<input name="textfield" type="text" value="<?php echo $position; ?>" size="40" />
<br />
<input name="textfield" type="text" value="<?php echo $practice; ?>" size="40" />
<br />
<input name="textfield" type="text" value="<?php echo $projname; ?>" size="40" />
<br />
<input name="textfield" type="text" value="<?php echo $projmgr; ?>" size="40" />
<br />
<input name="textfield" type="text" value="<?php echo $teamlead; ?>" size="40" />
<input type="submit" name="edit" value="Edit" />
<input type="submit" name="update" value="Update" />
</html>


but when i click my search button, i got this error


( ! ) Warning: mysql_query() expects parameter 1 to be string, resource given in C:\wamp\www\log-in\search.php on line 22
Call Stack
# Time Memory Function Location
1 0.0015 393280 {main}( ) ..\search.php:0
2 0.0182 402952 mysql_query ( ) ..\search.php:22

( ! ) Warning: mysql_num_rows() expects parameter 1 to be resource, null given in C:\wamp\www\log-in\search.php on line 23
Call Stack
# Time Memory Function Location
1 0.0015 393280 {main}( ) ..\search.php:0
2 0.0191 403200 mysql_num_rows ( ) ..\search.php:23

( ! ) Notice: Undefined variable: emp_number in C:\wamp\www\log-in\search.php on line 48 Call Stack #TimeMemoryFunctionLocation 10.0015393280{main}( )..\search.php:0 " size="40" />
( ! ) Notice: Undefined variable: fname in C:\wamp\www\log-in\search.php on line 50 Call Stack #TimeMemoryFunctionLocation 10.0015393280{main}( )..\search.php:0 " size="40" />
( ! ) Notice: Undefined variable: mname in C:\wamp\www\log-in\search.php on line 52 Call Stack #TimeMemoryFunctionLocation 10.0015393280{main}( )..\search.php:0 " size="40" />
( ! ) Notice: Undefined variable: lname in C:\wamp\www\log-in\search.php on line 54 Call Stack #TimeMemoryFunctionLocation 10.0015393280{main}( )..\search.php:0 " size="40" />
( ! ) Notice: Undefined variable: homeadd in C:\wamp\www\log-in\search.php on line 56 Call Stack #TimeMemoryFunctionLocation 10.0015393280{main}( )..\search.php:0 " size="40" />
( ! ) Notice: Undefined variable: emailadd in C:\wamp\www\log-in\search.php on line 58 Call Stack #TimeMemoryFunctionLocation 10.0015393280{main}( )..\search.php:0 " size="40" />
( ! ) Notice: Undefined variable: ofcnum in C:\wamp\www\log-in\search.php on line 60 Call Stack #TimeMemoryFunctionLocation 10.0015393280{main}( )..\search.php:0 " size="40" />
( ! ) Notice: Undefined variable: mobilenum in C:\wamp\www\log-in\search.php on line 62 Call Stack #TimeMemoryFunctionLocation 10.0015393280{main}( )..\search.php:0 " size="40" />
( ! ) Notice: Undefined variable: homenum in C:\wamp\www\log-in\search.php on line 64 Call Stack #TimeMemoryFunctionLocation 10.0015393280{main}( )..\search.php:0 " size="40" />
( ! ) Notice: Undefined variable: position in C:\wamp\www\log-in\search.php on line 66 Call Stack #TimeMemoryFunctionLocation 10.0015393280{main}( )..\search.php:0 " size="40" />
( ! ) Notice: Undefined variable: practice in C:\wamp\www\log-in\search.php on line 68 Call Stack #TimeMemoryFunctionLocation 10.0015393280{main}( )..\search.php:0 " size="40" />
( ! ) Notice: Undefined variable: projname in C:\wamp\www\log-in\search.php on line 70 Call Stack #TimeMemoryFunctionLocation 10.0015393280{main}( )..\search.php:0 " size="40" />
( ! ) Notice: Undefined variable: projmgr in C:\wamp\www\log-in\search.php on line 72 Call Stack #TimeMemoryFunctionLocation 10.0015393280{main}( )..\search.php:0 " size="40" />
( ! ) Notice: Undefined variable: teamlead in C:\wamp\www\log-in\search.php on line 74 Call Stack #TimeMemoryFunctionLocation 10.0015393280{main}( )..\search.php:0 " size="40" />

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daveismyname
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Joined: Sat Oct 15, 2011 3:21 am
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Re: SQL ERROR

Postby daveismyname » Thu May 16, 2013 3:35 am

sigridann wrote:$sql = mysql_query("select * from employee where fname like '%$term%'");
$result=mysql_query($sql);


your using mysql_query then again in the next line try this instead.

Code: Select all

$result = mysql_query("select * from employee where fname like '%$term%'");


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