Unable to catch dropdownlist value, error in SQL statement

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  • valeberry
    New Member
    • May 2007
    • 7
    #1

    Unable to catch dropdownlist value, error in SQL statement

    [PHP]index
    <?php
    include ("conn.php") ;
    require_once("t abs.php");
    ?>
    <html>
    <head>
    <?php tabs_header(); ?>
    </head>
    <body>
    <div style="width:60 0px;">
    <?php tabs_start(); ?>
    <?php tab( "Junior Colleges" ); ?>
    <form id="form1" name="form1" method="post" action="dd2.php ">
    <?PHP
    $query = 'SELECT DISTINCT schName FROM schools';
    //$query2='SELECT DISTINCT schLocation FROM schools';
    $result = mysql_query($qu ery) or die('Query failed: ' . mysql_error());
    echo "<select name=select value='' ><option>Plea se select</option>";
    while($nt=mysql _fetch_array($r esult)){//Array or records stored in $nt
    echo "<option value=$nt[schName]>$nt[schName]</option>";
    /* Option values are added by looping through the array */
    }
    echo "</select>";// Closing of list box

    echo '<input type="submit" value="Yes" />';



    ?></form>
    <?php tab( "PolyTechni c" ); ?>
    This is the second tab.
    <?php tabs_end(); ?>
    </div>
    </body>
    </html>
    [/PHP]

    [PHP]Display result
    <?PHP
    include ("conn.php") ;
    ?>
    <?PHP
    $schName=$_Get['select'];

    $query = "SELECT schName FROM schools where schName='$schNa me'";

    $numresults = mysql_query($qu ery) or die ("<center>Could n't execute query</center>");
    $row= mysql_fetch_arr ay($numresults) ;
    echo $numresults;
    echo $row;
    print $schName;
    echo '<br>';

    echo "<center><h1>Se arch results for: " . $schName ." </h1></center>";


    ?>
    [/PHP]

    Error Message when supposed to display result:
    Couldn't execute query
    "); $row= mysql_fetch_arr ay($numresults) ; echo $numresults; echo $row; print $schName; echo '
    '; echo "
    Search results for: " . $schName ."
    "; ?>


    Any1 knows what is wrong with my statement??
    Tags: None
    • ak1dnar
      Recognized Expert Top Contributor
      • Jan 2007
      • 1584
      #2
      [PHP]$schName=$_GET['select']; [/PHP]

        Comment

        • valeberry
          New Member
          • May 2007
          • 7
          #3
          Originally posted by ajaxrand
          [PHP]$schName=$_GET['select']; [/PHP]
          What shld i do to solve this error?
          Do i use $_Request to get the schName from drop downlist?

            Comment

            • pbmods
              Recognized Expert Expert
              • Apr 2007
              • 5821
              #4
              Originally posted by valeberry
              What shld i do to solve this error?
              Do i use $_Request to get the schName from drop downlist?
              Ajaxrand was pointing out that it s/b $_GET instead of $_Get.

              At this line:

              [PHP]
              $numresults = mysql_query($qu ery) or die ("<center>Could n't execute query</center>");
              [/PHP]

              Try this instead:
              [PHP]
              $numresults = mysql_query($qu ery) or die (mysql_error()) ;
              [/PHP]

              Also, try running your query manually in MySQL. Make sure you're actually getting results.

                Comment

                • ak1dnar
                  Recognized Expert Top Contributor
                  • Jan 2007
                  • 1584
                  #5
                  [PHP]$_Get [/PHP]is wrong
                  [PHP]$_GET[/PHP] is the correct way.

                    Comment

                    • ak1dnar
                      Recognized Expert Top Contributor
                      • Jan 2007
                      • 1584
                      #6
                      pbmods has posted before me. Thanks.:)
                      Valeberry did you solve the problem.

                        Comment

                        • valeberry
                          New Member
                          • May 2007
                          • 7
                          #7
                          Originally posted by ajaxrand
                          pbmods has posted before me. Thanks.:)
                          Valeberry did you solve the problem.
                          Yup. Thanks.
                          However, I'm still unable to display full name of the school from drop dwnlist. I'm getting from the database.
                          I hav insert the values.

                          [PHP]dd1
                          <?PHP
                          $query = 'SELECT DISTINCT schName FROM schools';
                          //$query2='SELECT DISTINCT schLocation FROM schools';
                          $result = mysql_query($qu ery) or die('Query failed: ' . mysql_error());
                          echo "<select name= 'select' value='' ><option>Plea se select</option>";
                          while($nt=mysql _fetch_array($r esult))
                          {
                          //Array or records stored in $nt
                          echo "<option value=$nt[schName]>$nt[schName]</option>";
                          /* Option values are added by looping through the array */
                          }
                          echo "</select>";// Closing of list box

                          echo '<input type="submit" value="Yes" />';



                          ?>


                          [/PHP]

                          Is there anything wrong with the logic with the follwin statement?
                          echo "<option value=$nt[schName]>$nt[schName]</option>";


                          [PHP]dd2
                          <?PHP
                          include ("conn.php") ;
                          ?>
                          <?PHP
                          $schName= $_GET['select'];
                          echo $_REQUEST[$schName];
                          $query = "SELECT schName FROM schools where schDesc='$schNa me'";

                          $numresults = mysql_query($qu ery) or die ('Query failed: ' . mysql_error());
                          $row= mysql_fetch_row ($numresults);
                          $row[0] =$schName ;


                          echo "<center><h1>Se arch results for: " . $schName ." </h1></center>";



                          ?>

                          [/PHP]

                          From the following codes:
                          $row= mysql_fetch_row ($numresults);
                          $row[0] =$schName ;

                          I can display Victoria instead of Victoria Junior College.

                          How shld I resolve the problem to display Victoria Junior College instead of Victoria?

                            Comment

                            • ak1dnar
                              Recognized Expert Top Contributor
                              • Jan 2007
                              • 1584
                              #8
                              You are still printing this variable
                              [PHP] $schName= $_GET['select']; [/PHP]

                              Not the Table value.Use like this.

                              [PHP]<?PHP
                              include ("conn.php") ;
                              $schName= $_GET['select'];
                              //echo $_REQUEST[$schName];
                              $query = "SELECT schName FROM schools where schDesc='$schNa me'";
                              $numresults = mysql_query($qu ery) or die ('Query failed: ' . mysql_error());
                              $row= mysql_fetch_row ($numresults);
                              //$row[0] =$schName ;
                              echo "<center><h1>Se arch results for: " . $row[0] ." </h1></center>";
                              ?> [/PHP]

                                Comment

                                • valeberry
                                  New Member
                                  • May 2007
                                  • 7
                                  #9
                                  Originally posted by ajaxrand
                                  You are still printing this variable
                                  [PHP] $schName= $_GET['select']; [/PHP]

                                  Not the Table value.Use like this.

                                  [PHP]<?PHP
                                  include ("conn.php") ;
                                  $schName= $_GET['select'];
                                  //echo $_REQUEST[$schName];
                                  $query = "SELECT schName FROM schools where schDesc='$schNa me'";
                                  $numresults = mysql_query($qu ery) or die ('Query failed: ' . mysql_error());
                                  $row= mysql_fetch_row ($numresults);
                                  //$row[0] =$schName ;
                                  echo "<center><h1>Se arch results for: " . $row[0] ." </h1></center>";
                                  ?> [/PHP]
                                  It did nt display ath. =(
                                  Output was a blank page.

                                    Comment

                                    • ak1dnar
                                      Recognized Expert Top Contributor
                                      • Jan 2007
                                      • 1584
                                      #10
                                      Originally posted by valeberry
                                      It did nt display ath. =(
                                      Output was a blank page.
                                      Is it working when you pass a value to $schName manually.

                                      [PHP]<?PHP
                                      include ("conn.php") ;
                                      //$schName= $_GET['select'];
                                      $schName= 'victoriya';
                                      //echo $_REQUEST[$schName];
                                      $query = "SELECT schName FROM schools where schDesc='$schNa me'";
                                      $numresults = mysql_query($qu ery) or die ('Query failed: ' . mysql_error());
                                      $row= mysql_fetch_row ($numresults);
                                      //$row[0] =$schName ;
                                      echo "<center><h1>Se arch results for: " . $row[0] ." </h1></center>";
                                      ?>[/PHP]

                                        Comment

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