Passing javascript variable to php with other html variable using post method

Question

I want to send variable rows to post2.php with other HTML form variable using POST or GET methods.

The below code gives an error:

Notice: Undefined index: row1 in C:\xampp\htdocs\PhpProject1\OtherUsableItems\post2.php on line 8

post1.php

 <html>
 <head>
 <SCRIPT language="javascript">
 function addRow(tableID) {
      var table = document.getElementById(tableID);

      var rowCount = table.rows.length;
      var row = table.insertRow(rowCount);

      var colCount = table.rows[0].cells.length;

      for(var i=0; i<colCount; i++) {
          var newcell = row.insertCell(i);

          newcell.innerHTML = table.rows[0].cells[i].innerHTML;
          //alert(newcell.childNodes);
          switch(newcell.childNodes[0].type) {
              case "text":
                      newcell.childNodes[0].value = "";
                      break;
              case "checkbox":
                      newcell.childNodes[0].checked = false;
                      break;
              case "select-one":
                      newcell.childNodes[0].selectedIndex = 0;
                      break;
          }
      }      
  }
  </script>
  <script>
  function count(tableId){
      var rows = document.getElementById(tableId).getElementsByTagName("TR").length;

     // window.location.href = "http://localhost/PhpProject1/OtherUsableItem  /post2.php?rows=" + rows ;
     // alert('Your table has ' + rows + ' rows.');
     $.post("post2.php", { 'row' : rows}, function(rows){alert('rows'+rows);});
  }
  </script>

  </head>
  <body>
  <form action="post2.php" method="post">

   <TABLE id="dataTable"  border="1">
    <TR>
        <TD> 1 </TD>
        <TD> <INPUT name="n1[]"type="text" /> </TD>
        <TD> <INPUT name="n2[]"type="text" /> </TD>
        <TD><SELECT name="country[]" type="select-one">
                <OPTION value="in">India</OPTION>
                <OPTION value="de">Germany</OPTION>
                <OPTION value="fr">France</OPTION>
                <OPTION value="us">United States</OPTION>
                <OPTION value="ch">Switzerland</OPTION>
            </SELECT></TD>
    </TR>
  </TABLE>
 <INPUT type="button" value="Add Row" onclick="addRow('dataTable');"/>
 <button id="bt" onclick="count('dataTable');">Submit</button>
 </form>

 </body>
 </html>

post2.php

<?php
    $n1 = $_POST['n1'];
    $n2 = $_POST['n2'];
    $country = $_POST['country'];
    echo $n1[0];
    echo $n2[0];
    echo $country[0];
    $count = $_POST['row1'];
    echo $count;
?>

Please write what does that echos writes, also you can try writing 'var_dump($_POST)' in post2.php to see full content of $_POST (usualy var_dump isnt formated so view source code or add <pre> tag) — kajacx
– kajacx, Commented Apr 22, 2013 at 19:20

George · Accepted Answer · 2013-04-22 20:22:21Z

1

Try changing to 'row' instead of 'row1'

$n1 = $_POST['n1'];
$n2 = $_POST['n2'];
$country = $_POST['country'];
echo $n1[0];
echo $n2[0];
echo $country[0];
$count = $_POST['row'];
echo $count;

In the future, use print_r to see the value of $_POST.

In addition to the above instructions, I would remove the 2nd <script> tag from post1.php and place the following code into the body at the start of the form:

<form action="post2.php" method="post" >
  <input id="rowNumber" type="hidden" name="row" value="1"/>

Also, add the following lines to function addRow:

    var rowNumber = document.getElementById('rowNumber');
    rowNumber.value = parseInt( rowNumber.value ) + 1;

edited Apr 22, 2013 at 20:22

answered Apr 22, 2013 at 19:17

George

3571 silver badge10 bronze badges

it's posting mistake by me . After changing row1 to row . i got Undefined index: row in C:\xampp\htdocs\PhpProject1\OtherUsableItems\post2.php on line 8

Abhishek
– Abhishek

04/22/2013 19:21:22
Commented Apr 22, 2013 at 19:21
in Notice: Undefined index: row in C:\xampp\htdocs\PhpProject1\OtherUsableItems\post2.php on line 8 Array ( [n1] => Array ( [0] => [1] => ) [n2] => Array ( [0] => [1] => ) [country] => Array ( [0] => in [1] => in ) ) 1

Abhishek
– Abhishek

04/22/2013 19:23:40
Commented Apr 22, 2013 at 19:23
If you are going to use jQuery in your implementation, be sure to actually include the jQuery script: <script src="//ajax.googleapis.com/ajax/libs/jquery/2.0.0/jquery.min.js"></script>

George
– George

04/22/2013 20:23:54
Commented Apr 22, 2013 at 20:23

Add a comment |

user1465831user1465831 · Accepted Answer · 2013-04-22 20:03:01Z

The problem is that you are not sending the correct Post value. check if this line :

var rows = document.getElementById(tableId).getElementsByTagName("TR").length;

it returns values something like: {name:'value',name2:'value2'}

after that you will be able to access via php using $_POST['name']...

and this line :

 $.post("post2.php", { 'row' : rows}, function(rows){alert('rows'+rows);});

replaced with:

 $.post("post2.php", rows, function(rows){alert('rows'+rows);});

else you will be accessing with $_POST['row']

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