Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given

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PHP: Warning: sort() expects parameter 1 to be array, resource given

here is my code: I see nothing that should be causing this...ideas?

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given.

 <?php

 include_once "mysql_connect.php";

 if ($_POST['parse_var'] == "contactform"){

 if(is_array($categories)) $whereCond = "in '".implode(",",$categories); else 
 $wherecond  = "= ".$categories;

       $dropdownValue = $_POST['dropdown'];

$dropdownValue = mysql_real_escape_string($dropdownValue);
$dropdownValue = eregi_replace("`", "", $dropdownValue);


$searchField= $_POST['searchinput'];
$searchField = mysql_real_escape_string($searchField);
$searchField = eregi_replace("`", "", $searchField);


if ($dropdownValue == "phone"){
$sql = mysql_query("SELECT * FROM pcparts WHERE phone='$searchField'");


    while($row1 = mysql_fetch_array($sql)){


        $arrayuserinfo[] = array(

            'phone' => $row1["phone"],
        'name' => $row1["name"],
        'city' => $row1["city"],
        'state' => $row1["state"],
        'address' => $row1["address"],
        'zip' => $row1["zip"],
    );
    };
                    for($i=0;$i < count($arrayuserinfo);$i++){

    $phone = $arrayuserinfo[$i]["phone"];
    $name = $arrayuserinfo[$i]["name"];
    $city = $arrayuserinfo[$i]["city"];
    $state = $arrayuserinfo[$i]["state"];
    $address = $arrayuserinfo[$i]["address"];
    $zip = $arrayuserinfo[$i]["zip"];
            echo"<table width='400' border='1' cellpadding='3'>
            <tr>
            <td>phone</td>
            <td>name</td>
            <td>address</td>
            <td>city</td>
            <td>state</td>
            <td>zip</td>
            </tr>
             <tr>
            <td>$phone</td>
            <td>$name</td>
             <td>$address</td>
             <td>$city</td>
            <td>$state</td>
           <td>$zip</td>
       </tr>
        </table><br />
        ";
        }
}

else if ($dropdownValue == "name"){
$sql = mysql_query("SELECT * FROM pcparts WHERE name='$searchField'");


    while($row1 = mysql_fetch_array($sql)){

        $arrayuserinfo[] = array(

        'phone' => $row1["phone"],
        'name' => $row1["name"],
        'city' => $row1["city"],
        'state' => $row1["state"],
        'address' => $row1["address"],
        'zip' => $row1["zip"],
    );
    };

        for($i=0;$i < count($arrayuserinfo);$i++){

    $phone = $arrayuserinfo[$i]["phone"];
    $name = $arrayuserinfo[$i]["name"];
    $city = $arrayuserinfo[$i]["city"];
    $state = $arrayuserinfo[$i]["state"];
    $address = $arrayuserinfo[$i]["address"];
    $zip = $arrayuserinfo[$i]["zip"];

    echo"<table width='400' border='1' cellpadding='3'>
        <tr>
        <td>phone</td>
        <td>name</td>
        <td>address</td>
         <td>city</td>
        <td>state</td>
        <td>zip</td>
        </tr>
         <tr>
        <td>$phone</td>
        <td>$name</td>
       <td>$address</td>
        <td>$city</td>
        <td>$state</td>
        <td>$zip</td>
       </tr>
       </table><br />
        ";
        }
}

else if ($dropdownValue == "city"){
    $sql = mysql_query("SELECT * FROM pcparts WHERE city='$searchField'");


    while($row1 = mysql_fetch_array($sql)){

        $arrayuserinfo[] = array(

        'phone' => $row1["phone"],
        'name' => $row1["name"],
        'city' => $row1["city"],
        'state' => $row1["state"],
        'address' => $row1["address"],
        'zip' => $row1["zip"],
    );
    };

        for($i=0;$i < count($arrayuserinfo);$i++){

    $phone = $arrayuserinfo[$i]["phone"];
    $name = $arrayuserinfo[$i]["name"];
    $city = $arrayuserinfo[$i]["city"];
    $state = $arrayuserinfo[$i]["state"];
    $address = $arrayuserinfo[$i]["address"];
    $zip = $arrayuserinfo[$i]["zip"];

    echo"<table width='400' border='1' cellpadding='3'>
     <tr>
     <td>phone</td>
     <td>name</td>
     <td>address</td>
     <td>city</td>
     <td>state</td>
     <td>zip</td>
     </tr>
       <tr>
    <td>$phone</td>
    <td>$name</td>
    <td>$address</td>
    <td>$city</td>
   <td>$state</td>
    <td>$zip</td>
    </tr>
    </table><br />
     ";
        }
}

      }
      ?>




     <html>
     <body>
      <h2>Customers Database Search</h2>

    <form action="file1.php" method="POST">
    <input type='hidden' name='parse_var' value='contactform'>
    Search fields:
 <select name="dropdown" id="dropdown" >
 <option value="<?php print "$dropdownValue"; ?>"><?php print "$dropdownValue"; ?></option>
  <option value="phone">phone</option>
  <option value="name">name</option>
  <option value="city">city</option>
</select>
 Search item:<input type="text" name='searchinput' id="searchinput" value="
 <?php print "$searchField"; ?>" size='20'><br><br>
 <input type="submit" name="button" id="button" value="Search" />
 </form>
 <br />
  <br />
 </body>
  </html>

mysql_connect:

   <?php

    $db_host = "localhost";
    $db_username = "user";

     $db_pass = "test1234";

    $db_name = "pcparts";

    @mysql_connect("$db_host","$db_username","$db_pass") or die ("Could not connect to MySQL");
    @mysql_select_db("$db_name") or die ("No database");

    ?>

if you find something i need to add, can you please tell me exactly where to put it?

Why are you ignoring the potential errors raised by `mysql_connect` and others? — Mat, Jul 17 '11 at 12:47
Your query failed and therefore it did not return anything that function which gives error can work with. That easy it is. You need to deal with the failures that might happen when code relies on functions which might fail. — hakre, Jul 17 '11 at 12:49

score 6 · Answer 1 · answered Jul 17 '11 at 12:48

6

Your query failed, and you forgot to use mysql_error() to find out why.

answered Jul 17 '11 at 12:48

Ignacio Vazquez-Abrams

740,318
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score 2 · Answer 2 · answered Jul 17 '11 at 12:48

2

mysql_query returns FALSE when an error occurs. Make sure you deal with that situation before you try to use the $sql variable.

answered Jul 17 '11 at 12:48

Mat

195,986
40
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396

Michael Berkowski · Answer 3 · 2011-07-17T13:50:50.853

One of your queries is failing, or you never had a valid connection to begin with.

Remove the @ from the mysql_connect() and mysql_select_db() statements. Then after each call to mysql_query(), you need to check if it succeeded:

$sql = mysql_query("SELECT * FROM pcparts WHERE city='$searchField'");
if ($sql) {
  // query was valid -- fetch results.
}
else {
  // something went wrong.
  echo mysql_error();
}

Just a tip - you have piles and piles of unnecessary code copy/pasted. This whole block:

while($row1 = mysql_fetch_array($sql)){


        $arrayuserinfo[] = array(

            'phone' => $row1["phone"],
        'name' => $row1["name"],
        'city' => $row1["city"],
        'state' => $row1["state"],
        'address' => $row1["address"],
        'zip' => $row1["zip"],
    );
    };
                    for($i=0;$i < count($arrayuserinfo);$i++){

    $phone = $arrayuserinfo[$i]["phone"];
    $name = $arrayuserinfo[$i]["name"];
    $city = $arrayuserinfo[$i]["city"];
    $state = $arrayuserinfo[$i]["state"];
    $address = $arrayuserinfo[$i]["address"];
    $zip = $arrayuserinfo[$i]["zip"];
            echo"<table width='400' border='1' cellpadding='3'>
            <tr>
            <td>phone</td>
            <td>name</td>
            <td>address</td>
            <td>city</td>
            <td>state</td>
            <td>zip</td>
            </tr>
             <tr>
            <td>$phone</td>
            <td>$name</td>
             <td>$address</td>
             <td>$city</td>
            <td>$state</td>
           <td>$zip</td>
       </tr>
        </table><br />
        ";
        }

Can be simplified down to just this:

while($row1 = mysql_fetch_array($sql)) {
  echo"<table width='400' border='1' cellpadding='3'>
      <tr>
            <td>phone</td>
            <td>name</td>
            <td>address</td>
            <td>city</td>
            <td>state</td>
            <td>zip</td>
            </tr>
             <tr>
            <td>{$row1['phone']}</td>
            <td>{$row1['name']}</td>
             <td>{$row1['address']}</td>
             <td>{$row1['city']}</td>
            <td>{$row1['state']}</td>
           <td>{$row1['zip']}</td>
      </tr>
    </table><br />";
}

Another tip: Every time you enclose a variable in quotes like these:

mysql_select_db("$db_name")
<?php print "$searchField"; ?>

It's redundant and unnecessary. You just need

mysql_select_db($db_name)
<?php print $searchField; ?>

ok, thanks. ran that and got this error: the table 'mytablename.mytablename' doesn't exist. I know it does so any idea what that problem could be? — gregg, Jul 17 '11 at 12:57
no...and that my be my problem i'm realizing. My database is named pcparts and my table is named people — gregg, Jul 17 '11 at 13:35
yes, i've fixed that but now it seems for some reason my submit button is no longer working. If its not one thing, its another right? Any more suggestions? — gregg, Jul 17 '11 at 13:46
@gregg Probably because I left off the `
` in my optimization and broke your html. Fixed above. — Michael Berkowski, Jul 17 '11 at 13:51

score 0 · Answer 4 · answered Jul 17 '11 at 12:49

Knowing which of the 3 mysql_fetch_array() calls is causing the warning would be helpful, as would knowing what value of $searchField is being passed, but looking at your code it suggests the SQL query you're forming is invalid. A failing query will return false instead of a mysql resource.

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given

4 Answers4