Parse error: syntax error, unexpected T_ELSE

[mikebyrne]

Im trying to code a searchfield in my site but im getting the error:

 

 

Parse error: syntax error, unexpected T_ELSE in C:\xampp\htdocs\MainPage\login.php on line 138

 

line 138 is: } else {

 

My code so far is:

 

<table width="100%" cellpadding="0" cellspacing="0" border="0" class="b1sw2" style="background:url(Pictures/searchbox_bg.gif) no-repeat #003;">
<col width="1%">
<col width="99%">

<tr>
<td> </td>
<td>
<label for="searchfield" class="t10nw"></label><br /><br /><br /><br /><br /><br />
<?PHP
include("adminconnect.php");
$tbl_name = "product";

$cat = mysql_real_escape_string($_POST['cat']);
$input =
mysql_real_escape_string($_POST['searchfield']);

$query = "SELECT * FROM $tbl_name WHERE $cat LIKE '%$input%'";
$result = mysql_query($result);

while($row = mysql_fetch_array &&
mysql_num_rows($result) > 0) {
$product = $row['product'];
echo "".$product." was found. ";
} else {
echo "No search results found";
}
?>
<select name="cat" onChange="setAction(this.options[this.selectedIndex].value);"> 
<option value="cd" selected="selected" >CDs</option>
<option value="dvd" >DVDs</option>
<option value="game" >Games</option>
</select>
<form name="ex2" method="get" action="/searchresults_cd.php">

<input type="text" name="searchfield" size="22" maxlength="40" id="srchdrop" />
<input type="submit" value="Search Now! »" id="gosrch" />

</td></tr>
</form>
</table>

[conker87]

You can't have an else after a while.

[mikebyrne]

How can I get round that??

[conker87]

try:

while($row = mysql_fetch_array) {
if *mysql_num_rows($result) > 0)
{
$product = $row['product'];
echo "".$product." was found. ";
} else {
echo "No search results found";
} }

[kenrbnsn]

Do this instead:

<?php
$result = mysql_query($result);
if (mysql_num_rows($result) > 0)) {
     while($row = mysql_fetch_array) {
         $product = $row['product'];
         echo "$product was found<br>";
     }
} else {
    echo "No search results found";
}
?>

 

Ken

[mikebyrne]

Getting the error

 

 

Parse error: syntax error, unexpected ')' in C:\xampp\htdocs\MainPage\login.php on line 134

 

Line 134 is:

 

if (mysql_num_rows($result) > 0)) {

[dmccabe]

delete one of the ) after the 0 you have 1 too many

[mikebyrne]

Thanks!

 

My code now posts to searchresults.php. What needs to be here to display the resuts. Is it just a matter of echoing $product??

 

 

[conker87]

Do this instead:

<?php
$result = mysql_query($result);
if (mysql_num_rows($result) > 0)) {
     while($row = mysql_fetch_array) {
         $product = $row['product'];
         echo "$product was found<br>";
     }
} else {
    echo "No search results found";
}
?>

 

Ken

Yeah that lol

 

<-- fail

[AndyB]

Yeah that lol

 

<-- fail

 

"Fail" is useless to us in terms of helping with your problem.  Please describe what happens, why you consider it a failure, what error messages if any are displayed, etc. etc.

[mikebyrne]

I DIDNT MENTION "FAIL" (its not in my vocab!) lol

 

At present ive coded a serchfield on my login page which should pass the results to searchresults.php. I presume i need to use the $_GET [Result];

 

I'm just not how to code it to see whats getting passed over

[conker87]

Lol, sorry, I actually meant I failed. As in I failed at that piece of code. Not the code failed.

 

Just to clear that up ^^

[mikebyrne]

I know, no problem.

 

Any idea how I can ger $result to show on show on searchresults.php??