[SOLVED] Query not returning any results

[greencoin]

HI! was getting another error but read about the mysql functions and since modified my code. Now I'm getting "No Results Found" which is one of the "else" returns as you'll see.

 

<?php 

// db connect stuff goes here 

$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)";
$sql = "SELECT * FROM GC_Tracker WHERE $name2q";
if ($result = mysql_query($sql)) {
      if (mysql_num_rows($result)) {
        $i = 0;
        echo "<TABLE width=\"850\" border=\"1\" cellpadding=\"0\" cellspacing=\"0\">\n";
        echo "<TR bgcolor=\"ccffcc\"><TD>ID</TD><TD width=130>Customer</TD><TD width=75>Area</TD><TD width=75>City</TD><TD width=100>Phone</TD><TD width=75>Amount</TD><TD>Units</TD></TR>\n";
        while ($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"; // this is what it's spitting out
      }
    } else {
      echo "Query failed<br />" . $sql . "<br />" . mysql_error();
  }
?>

 

Could someone point me in the right direction to debug or point out the flaw in my code. The database is loaded so I should've had a result based on the input.

[chocopi]

i think it could be

 

if ($result = mysql_query($sql)) {

 

but $result isnt defined anywhere and if you comparing shouldnt you use ==

 

~ Chocopi

[greencoin]

I recycled this code from a working page on the same website that queries MySQL for all records of a specific field. I don't think I'll need the == as I'm not comparing.

 

I created an array that posts all row data with corresponding field data (supposed to). ~Rich

[Wildbug]

Print out the $sql query that's being built.

[greencoin]

"Parse error: parse error, unexpected T_IF in http://./././track_area_results.php on line 16"

 

<?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)";
$sql = "SELECT * FROM GC_Tracker WHERE $name2q";
die ($sql)
if ($result = mysql_query($sql)) {                               // This is line 16
      if (mysql_num_rows($result)) {
        $i = 0;
        echo "<TABLE width=\"850\" border=\"1\" cellpadding=\"0\" cellspacing=\"0\">\n";
        echo "<TR bgcolor=\"ccffcc\"><TD>ID</TD><TD width=130>Customer</TD><TD width=75>Area</TD><TD width=75>City</TD><TD width=100>Phone</TD><TD width=75>Amount</TD><TD>Units</TD></TR>\n";
        while ($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();
  }
?>

 

Thanks Wildbug ~Rich

[Wildbug]

You need a semi-colon after the "die ($sql)" statement; the parser is running into the if() and isn't expecting it.

[greencoin]

ack! I keep forgettin my semicolons... here's what you're lookin for;

 

SELECT * FROM GC_Tracker WHERE ("area" LIKE '%ak%' OR "area" LIKE '%LA%' )

 

whatcha think? ~Rich

[greencoin]

*Bump* ~Rich

[teng84]

SELECT * FROM GC_Tracker WHERE ("area" LIKE '%ak%' OR "area" LIKE '%LA%' )

 

^^ does that thing work???

[greencoin]

Ahhh... you tell me. I believe the syntax is correct and the field + variables are correct too. Am going to bed now. Ttyl ~Rich

[greencoin]

Ok back - anyone have any suggestions? ~Rich

[Wildbug]

Yeah, take the double quotes off the "area"; you're specifying the string "area" by enclosing it in quotes; you want the column area.  If you really, really want to quote the column, use backticks (`area`), not double or single quotes.

[greencoin]

I am humbled and ashamed... the original code used bar ticks but I changed them trying to figure out the problem.

 

Wildbug.. You da schiz-nit! Thanks for your help, again. ~Rich