How to make autopost back for 2 buttons on same page using php

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  • vadakattunaveen
    New Member
    • Jan 2013
    • 3

    How to make autopost back for 2 buttons on same page using php

    In php page i have 2 buttons 1st button is for retrieving data from db and 2nd button also is for retrieving data from db in execution when i pressed 1st button its showing data and at the same time when i pressed 2nd button its showing data but same time the page losing 1st button data .i want to be displayed both button data at same time can any one help me please????????? ????????????








    Code:
    <html>
    <head>
    <title>
    Search details Form
    </title>
    </head>
    <body>
    <form method = "POST" action = "search2.php">
    Dealer ID<input type = "text" id = "DealerId" name = "DealerId">
    <input type = "submit" name = "submit1" value = "Search1" >
    <?php
    if(isset($_POST['submit1']))
    {
    $name = $_POST['DealerId'];
    if($name != "")
    {
    mysql_connect("localhost", "root", "");
    mysql_select_db("rajesh");
    $sql = "select * from dealerlist where DealerID like '%".mysql_real_escape_string($name)."%'";
    $result = mysql_query($sql);
    $numRows = mysql_num_rows($result);
    if(empty($numRows))
    {
    	echo "Record are not exits";
    }
    
    else{
    while($row = mysql_fetch_array($result))
    {
    ?>
    <table>
    <tr><td>Dealer ID<td><input type = "text" name = "Name" value = "<?php echo $row['DealerID']; ?>">
    <tr><td>Dealer Name<td><input type = "text" name = "Age" value = "<?php echo $row['DealerName']; ?>">
    <tr><td>Address<td><input type = "text" name = "Sex" value = "<?php echo $row['Street']; ?>"></tr>
    </table><br>
    ItemNo<input type = "text" name = "ItemNo">
    <input type = "submit" name = "submit2" value = "search2">
    <?php
    }
    }
    }
    }
    if(isset($_POST['submit2']))
    {
    $name1 = $_POST['ItemNo'];
    if($name1 != "")
    {
    mysql_connect("localhost", "root", "");
    mysql_select_db("rajesh");
    $sql1 = "select * from itemlist where ItemNO  like '%".mysql_real_escape_string($name1)."%'";
    $result1 = mysql_query($sql1);
    $numRows1 = mysql_num_rows($result1);
    if(empty($numRows1))
    {
    	echo "Record are not exits";
    }
    else{
    while($row1 = mysql_fetch_array($result1))
    {
    ?>
    <table>
    <tr><td>Item No<td><input type = "text" name = "ItemNo" value = "<?php echo $row1['ItemNo']; ?>">
    <tr><td>Item Name<td><input type = "text" name = "ItemName" value = "<?php echo $row1['ItemName']; ?>">
    </table>
    <?php
    }
    }
    }
    }
    ?>
  • vadakattunaveen
    New Member
    • Jan 2013
    • 3

    #2
    Code:
    [LEFT][CENTER]<head>
    <title>
    Search details Form
    </title>
    </head>
    <body>
    <form method = "POST" action = "search2.php">
    Dealer ID<input type = "text" id = "DealerId" name = "DealerId">
    <input type = "submit" name = "submit1" value = "Search1" >
    <?php
    if(isset($_POST['submit1']))
    {
    $name = $_POST['DealerId'];
    if($name != "")
    {
    mysql_connect("localhost", "root", "");
    mysql_select_db("rajesh");
    $sql = "select * from dealerlist where DealerID like '%".mysql_real_escape_string($name)."%'";
    $result = mysql_query($sql);
    $numRows = mysql_num_rows($result);
    if(empty($numRows))
    {
    	echo "Record are not exits";
    }
    
    else{
    while($row = mysql_fetch_array($result))
    {
    ?>
    <table>
    <tr><td>Dealer ID<td><input type = "text" name = "Name" value = "<?php echo $row['DealerID']; ?>">
    <tr><td>Dealer Name<td><input type = "text" name = "Age" value = "<?php echo $row['DealerName']; ?>">
    <tr><td>Address<td><input type = "text" name = "Sex" value = "<?php echo $row['Street']; ?>"></tr>
    </table><br>
    ItemNo<input type = "text" name = "ItemNo">
    <input type = "submit" name = "submit2" value = "search2">
    <?php
    }
    }
    }
    }
    if(isset($_POST['submit2']))
    {
    $name1 = $_POST['ItemNo'];
    if($name1 != "")
    {
    mysql_connect("localhost", "root", "");
    mysql_select_db("rajesh");
    $sql1 = "select * from itemlist where ItemNO  like '%".mysql_real_escape_string($name1)."%'";
    $result1 = mysql_query($sql1);
    $numRows1 = mysql_num_rows($result1);
    if(empty($numRows1))
    {
    	echo "Record are not exits";
    }
    else{
    while($row1 = mysql_fetch_array($result1))
    {
    ?>
    <table>
    <tr><td>Item No<td><input type = "text" name = "ItemNo" value = "<?php echo $row1['ItemNo']; ?>">
    <tr><td>Item Name<td><input type = "text" name = "ItemName" value = "<?php echo $row1['ItemName']; ?>">
    </table>
    <?php
    }
    }
    }
    }
    ?>[LEFT][/LEFT][/CENTER][/LEFT]
    <html>

    Comment

    • Rabbit
      Recognized Expert MVP
      • Jan 2007
      • 12517

      #3
      If you want to run both, don't check to see which button was clicked.

      Comment

      • vadakattunaveen
        New Member
        • Jan 2013
        • 3

        #4
        First of all thanks mr.rabbit for responding to my question i have given u code and wts the problem pls follow of my code and reply once again .actually i want 2 buttons results to be there in page.

        Comment

        • Rabbit
          Recognized Expert MVP
          • Jan 2007
          • 12517

          #5
          Yes, I know. I answered your question.

          Comment

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