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Editorial Team
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Asked: 2026-05-31T13:14:42+00:00

I try to show Mysql result as table inside <div> after click submit button,

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I try to show Mysql result as table inside <div> after click submit button, but it just only show <table></table>. No problem found during posting value to process page.
So far, my script is like: 

<form id="myform">
....................
<button id="input" type="button" class="ui-state-default ui-corner-all"><span>Submit </span></button>
<input name="action" value="openreport" type="hidden">
</form>
<div id="show"></div>

$("#submit").click(function(){
      var params=$("#myform").serialize();
      $.ajax({
              type:"post",
              url:"go.php",
              data:params,
              cache :false,
              async :false,
              success : function(result) {
                   $('#show').replaceWith(result);
                   }
              });
      });

page go.php: 

     <?php

         //CONNECT TO DATABASE
         $dbc=mysql_connect(_SRV, _ACCID, _PWD) or die(_ERROR15.": ".mysql_error());
         mysql_select_db("qdbase") or die(_ERROR17.": ".mysql_error());

         switch (postVar('action')) {
                 case 'openreport':
                         openreport(postVar('model'),postVar('line'),postVar('lot_no'));
                         break;
                         }

         function openreport($model,$line,$lot_no){
                 $Model = mysql_real_escape_string($model);
                 $Line = mysql_real_escape_string($line);
                 $Lot = mysql_real_escape_string($lot_no);

                 $group=" GROUP BY DATE ";

                 $sql="SELECT Range_sampling,DATE(Inspection_datetime) AS DATE FROM
     inspection_report WHERE Model LIKE '".$Model."'"; 
                     $sql.="AND Line LIKE '".$Line."' AND Lot_no LIKE '".$Lot."'".$group;
                     $result=mysql_query($sql) or die(_ERROR26.": ".mysql_error());

                     echo "<table border='1'>


 <tr>
     <th>Firstname</th>
     <th>Lastname</th>
     </tr>";
     while($row = mysql_fetch_array($result))
       {
     echo("<tr><td>$row[0]</td><td>$row[1]</td></tr>");
     //  echo "<tr>";
     //  echo "<td>" . $row['Range_sampling'] . "</td>";
     //  echo "<td>" . $row['DATE'] . "</td>";
     //  echo "</tr>";
       }
     echo "</table>";

     mysql_close($dbc);
         }
         ?>

I have no idea because I’m not really understand how to show mysql result as a html table.

only show table header

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1 Answer

  1. Editorial Team

    The AJAX call is made only if a button is clicked that has an id of “submit”; however, your submit button has an id of “input”. Try changing the line that reads:

    $("#submit").click(function(){

    to

    $("#input").click(function(){

    As Yun suggested, to keep <div id="show">...</div> wrapped around the table, change

    $('#show').replaceWith(result);

    with

    $('#show').html(result);

    I don’t know if this will help or not, but it’s the only thing that stands out to me.

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