[SOLVED] *modified * Query Failed - Resource id #2, error in syntax !!! Officially lost

[greencoin]

Here's what I get

"Query failed

Resource id #2

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Resource id #2' at line 1"

 

here's the code;

<?php
// db connect stuff
$name2 = $_POST['state'];
for($name2array=0; $name2array < sizeof($name2); $name2array++)
{
if($name2array < (sizeof($name2)-1)) { $name2_cond = " OR "; }
else { $name2_cond = ""; }
$name2q = $name2q."\"area\" LIKE'".$name2[$name2array]."'$name2_cond";
}
$name2q = "($name2q)";
$query = "SELECT * FROM GC_Tracker WHERE $name2q";
die($query);
$sql = mysql_query($query) or die(mysql_error());
   
    if ($result = mysql_query($sql)) {
      if (mysql_num_rows($result)) {
        $i = 0;
        echo "<TABLE width=\"850\" cellpadding=\"3\" cellspacing=\"1\" bgcolor=cccccc>\n";
        echo "<TR bgcolor=\"ccffcc\"><TD>ID</TD><TD width=130>Customer</TD><TD width=75>Area</TD><TD width=75>City</TD><TD width=100>Phone</TD><TD width=75>Amount</TD><TD>Units</TD></TR>\n";
        while ($row = mysql_fetch_array($result)) {
          if ($i % 2) {
            echo "<TR bgcolor=\"ccffcc\">\n";
          } else {
            echo "<TR bgcolor=\"white\">\n";
          }
          echo "<TD>".$row['cid']."</TD><TD>".$row['customer']."</TD><TD>".$row['area']."</TD><TD>".$row['city']."</TD>
	  <TD>".$row['phone']."</TD><TD>".$row['amount']."</TD>
	  <TD>".$row['prod1'].", ".$row['prod2'].", ".$row['prod3'].", ".$row['prod4'].", ".$row['prod5'].", ".$row['prod6'].", ".$row['prod7'].", ".$row['prod8'].", ".$row['prod9'].", ".$row['prod10']."</TD>\n";
          echo "</TR>\n";
          $i++;
        }
        echo "</TABLE>\n";
	echo "<br>\n";
	echo "<input type=Button name=printit value=print onclick=javascript:window.print();>\n";
      } else {
        echo "No results found";
      }
    } else {
      echo "Query failed<br />" . $sql . "<br />" . mysql_error();
    }
//  } else {
//   echo "form not submitted";
//  }

?>

 

The variables in the query are passing correctly. Using die($query) the string looks like; SELECT * FROM GC_Tracker WHERE ("area" LIKE'ak')

 

Other than there is no space in between LIKE and the variable 'ak', I don't see anything wrong with the SELECT line. Anyone? Thanks in advance ~Rich

[bbaker]

<?php
$name2 = $_POST['state'];
for($name2array=0; $name2array < sizeof($name2); $name2array++)
{
if($name2array < (sizeof($name2)-1)) { $name2_cond = " OR "; }
else { $name2_cond = ""; }
//$name2q = $name2q."/"area/" LIKE'".$name2[$name2array]."'$name2_cond"; // this is my variable that is on the SELECT line
$name2q = $name2q."area LIKE '".$name2[$name2array]."'$name2_cond"; // this is my variable that is on the SELECT line
}
//$name2q = "($name2q)";
$query = "SELECT * FROM GC_Tracker WHERE $name2q"; // I think this is considered line 1
$sql = mysql_query($query) or die(mysql_error());?>

 

should output as: SELECT * FROM GC_Tracker WHERE area LIKE 'ak'

 

however....if you're looking for field containing 'ak' you'll want to change it to

 

SELECT * FROM GC_Tracker WHERE area LIKE '%ak%'

[greencoin]

bbaker: I tried your correction and die($query) gave me;

SELECT * FROM GC_Tracker WHERE

 

When I uncommented $name2q = "($name2q)"; it gave me;

SELECT * FROM GC_Tacker WHERE ()

 

I DO need to find a field containing the variable so how would I pass the '%...%' to the SELECT statement?

 

I forgot to mention the form that passes the $name2 variable is a multiple checkbox form that will pass more than one variable at a time. Thanks ~Rich

[greencoin]

Trying to debug, I changed the code after finding some other people who posted the same errors. Now I have this code;

<?php
$name2 = $_POST['state'];
for($name2array=0; $name2array < sizeof($name2); $name2array++)
{
if($name2array < (sizeof($name2)-1)) { $name2_cond = " OR "; }
else { $name2_cond = ""; }
$name2q = $name2q."\"area\" LIKE '%".$name2[$name2array]."%' $name2_cond";

}
$name2q = "($name2q)";
$query = "SELECT * FROM GC_Tracker WHERE $name2q";
$sql = mysql_query($query) or die(mysql_error());

Print "<table width=860 border=1 cellpadding=0 cellspacing=0 class=table_results>"; 
Print "<tr bgcolor=ccffcc>"; 
Print "<td width=35>Order</td>";
Print "<td width=100>Customer</td>";
Print "<td width=75>Phone</td>";
Print "<td width=50>Area</td>";
Print "<td width=75>City</td>";
Print "<td width=70>Amount</td>";
Print "<td width=100>Products</td>";
Print "<td width=20>Created On</td></tr>";
while($info = mysql_fetch_array( $sql )) 
{ 

Print "<tr bgcolor=ffffff>"; 
Print "<td>".$info['cid'] . "</td> ";
Print "<td>".$info['customer'] . "</td> "; 
Print "<td>".$info['phone'] . " </td>";
Print "<td>".$info['area'] . " </td>";
Print "<td>".$info['city'] . "</td> "; 
Print "<td>".$info['amount'] . " </td>";
Print "<td>".$info['prod1'].  " </td>";
$var = $info['time'];
list($date,$time) = explode(' ',$var);
Print "<td>".$date . " </td></tr>"; 
} 
echo $sql; // for debugging only to check my variable

Print "</table>";
Print "<br>\n";
Print "<input type=Button name=printit value=print onclick=javascript:window.print();>\n"; 
?> 

 

It no longer kicks out an error but the beginings of a table, with blank rows of course. I echoed $sql and it returns "Resource #2". die ($query) gives me; SELECT * FROM GC_Tracker WHERE ("area" LIKE '%LA%' )

 

I'm using MySQL 4.1 on godaddy servers... ~Rich

[greencoin]

ok - modified the code and now getting a different error. will close this and open another thread based on the new problem. Thanks for the help as always. ~Rich