T_Logical_OR Parse error. Help is appreciated.

[todayme]

I have this error below, anyone got any ideas, I am sure it is a simple syntax error.

 

Parse error: syntax error, unexpected T_LOGICAL_OR in /home/admin/domains/sold.au.com/public_html/getlocation.php on line 16

 

 

<?

//set the variables as required

$dbhost = "localhost";

$dbuser = "admin_sex";

$dbpass = "sex";

$dbname = "admin_sex";

 

 

 

$db = mysql_pconnect($dbhost,$dbuser,$dbpass);

mysql_select_db($dbname) or die(mysql_error());

 

 

 

$data = mysql_query("SELECT * FROM _Industries WHERE Industry = '$dbIndustry'");

or die(mysql_error());

 

$info = mysql_fetch_array( $data );

 

$locationtype = $info['Location_Type'];

 

?>

<select name="Location"

><?php

 

  $query=mysql_query("("SELECT '$dbState' FROM '$locationtype'");

  while ($fetch=mysql_fetch_assoc($query)) {

    echo '<option value="'.$fetch['$dbState'].'">'.$fetch['$dbState'].'</option>';

  }

mysql_close($db);?></select>

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

[r-it]

the extra brackets in your query, the 2nd query

[monk.e.boy]

  $query=mysql_query("("SELECT '$dbState' FROM '$locationtype'");

 

This line looks like it has a syntax error

 

  $query=mysql_query("SELECT '$dbState' FROM '$locationtype'");

 

monk.e.boy

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

[todayme]

I found that one, and removed it but it still has the same error.  Anyone else got a idea!  Have I passed those variables into the second query correctly?

[redarrow]

<?php
//set the variables as required
$dbhost = "localhost"; 
$dbuser = "admin_sex"; 
$dbpass = "sex"; 
$dbname = "admin_sex"; 

$db = mysql_pconnect($dbhost,$dbuser,$dbpass); 
mysql_select_db($dbname) or die(mysql_error());

$data ="SELECT * FROM _Industries WHERE Industry = '$dbIndustry' ";

$result1=mysql_query($data)or die("mysql_error()");

$info = mysql_fetch_array( $result1); 

$locationtype = $info['Location_Type']; 

?>
<select name="Location" >

<?php

  $query="SELECT '$dbState' FROM '$locationtype'";
  $result2=mysql_query($query) or dir("mysql_error()");
  
  while ($fetch=mysql_fetch_assoc($result2)) {
    echo '<option value="'.$fetch['$dbState'].'">'.$fetch['$dbState'].'</option>';
  }
mysql_close($db);?></select>

[todayme]

Geting closer now I have this error after trying your code.

 

I am looking looking looking ha ha.

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/admin/domains/sold.au.com/public_html/getlocation.php on line 19

[monk.e.boy]

  $result2=mysql_query($query) or dir("mysql_error()");

 

 

or die?

 

monk.e.boy

[redarrow]

what?

sorry corrected.

<?php
//set the variables as required
$dbhost = "localhost"; 
$dbuser = "admin_sex"; 
$dbpass = "sex"; 
$dbname = "admin_sex"; 

$db = mysql_pconnect($dbhost,$dbuser,$dbpass); 
mysql_select_db($dbname) or die(mysql_error());

$data ="SELECT * FROM _Industries WHERE Industry = '$dbIndustry' ";

$result1=mysql_query($data)or die("mysql_error()");

$info = mysql_fetch_array( $result1); 

$locationtype = $info['Location_Type']; 

?>
<select name="Location" >

<?php

 $query="SELECT '$dbState' FROM '$locationtype'";
 $result2=mysql_query($query) or die("mysql_error()");
 
 while ($fetch=mysql_fetch_assoc($result2)) {
   echo '<option value="'.$fetch['$dbState'].'">'.$fetch['$dbState'].'</option>';
 }
mysql_close($db);?></select>

[todayme]

No errors coming up but the code doesnt work !............Thanks heaps for trying I am grrrrrrrrrr frustrated and going for a smoke............

 

Be back in 5 minutes, anyone else got an idea?

[todayme]

Havent gone for smoke yet just tried the second last lot of code again and it executed with no errors but no data in the drop down box.............I am going to look at this further after I have a smoke ha ha

[redarrow]

$query="SELECT

 

 

'$dbState'  << this is wrong ok?

 

 

 

FROM '$locationtype'";

[todayme]

Why is it wrong cant i pass a variable there instead of its value statically ?

[kenrbnsn]

Please post your current code between

tags.

 

Ken

[todayme]

Here is the correct code incase anyone wants it.

 

<?php
//set the variables as required
$dbhost = "localhost"; 
$dbuser = "admin_sex"; 
$dbpass = "sex"; 
$dbname = "admin_sex"; 

$db = mysql_pconnect($dbhost,$dbuser,$dbpass); 
mysql_select_db($dbname) or die(mysql_error());

$data ="SELECT * FROM _Industries WHERE Industry = '$dbIndustry' ";

$result1=mysql_query($data)or die("mysql_error()");

$info = mysql_fetch_array( $result1); 

$locationtype = $info['Location_Type']; 




?>
<select name="Location" >

<?php

  $query=mysql_query("SELECT $dbState FROM $locationtype");
  while ($fetch=mysql_fetch_assoc($query)) {
    echo '<option value="'.$fetch[$dbState].'">'.$fetch[$dbState].'</option>';
  }
mysql_close($db);?></select>

[code/]