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plusnplus

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Posts posted by plusnplus

  1. Hi,

    I have very long query in mysql using php.

    in apache already set the time out very long, 1000000.

    but still the process stop just about 1-2 minute.

     

    is in mysql also have setting i should change the value of time-out?

     

    thanks for any reply/ help

  2. Hi..,

    between:

    1. put image file (jpg/ gif only) to mysql db use bolb data type.

    2 put the link\ picture location path to mysql db use varchar data type.

     

    picture file condition:

    1. estimated 1000-1200 images

    2. each picture is less then 10 kb, max 800 x 600px

     

    which one is best way to store image file to database and why?

     

    Thanks for any info/ reply.

     

     

  3. Hi,

     

    I have code:

     

    do-login.php

    <?

    include("function/db.php");

    $user_id=$_GET['user_id'];

    $user_pwd=$_GET['user_pwd'];

    $link_id=$mylink;

    $user1=auth_login($user_id,$user_pwd,$link_id);

    if($user1==0) // user or password is wrong

    {

    $display_string="wrong";

    echo $display_string;

     

    }

    else

    {

    $display_string="correct";

    echo $display_string;

    }

    ?>

     

    login.php

    <html>

    <body topmargin="0"><center>

    <script language="javascript" type="text/javascript">

    <!--

    //Browser Support Code

    function ajaxFunction(){

    var ajaxRequest;  // The variable that makes Ajax possible!

     

    try{

    // Opera 8.0+, Firefox, Safari

    ajaxRequest = new XMLHttpRequest();

    } catch (e){

    // Internet Explorer Browsers

    try{

    ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");

    } catch (e) {

    try{

    ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");

    } catch (e){

    // Something went wrong

    alert("Your browser broke!");

    return false;

    }

    }

    }

    // Create a function that will receive data sent from the server

    ajaxRequest.onreadystatechange = function(){

    if(ajaxRequest.readyState == 4){

    var ajaxDisplay = document.getElementById('ajaxDiv');

    ajaxDisplay.innerHTML = ajaxRequest.responseText;

    }

    }

    var user_pwd = document.getElementById('user_pwd').value;

    var user_id = document.getElementById('user_id').value;

    var queryString = "?user_id=" + user_id + "&user_pwd=" + user_pwd;

    ajaxRequest.open("GET", "do-login.php" + queryString, true);

    ajaxRequest.send(null);

    }

    //-->

    </script>

     

    <form name='myForm'>

    <table border="0" bgcolor="E2EAFB"  cellspacing="2" cellpadding="2" width="300" class="Body">

        <tr valign="top">

    <td></td><td width=100>User name</td><td width=20>:</td>

    <td width=180><input type=text class="Body" size=18 id="user_id"></td>

    </tr>

    <tr valign="top">

    <td></td><td >Password</td><td>:</td>

    <td><input type=password class="Body" size=18 id="user_pwd"></td>

    </tr>

    <td colspan="4" align=center height=50><input type='button' onclick='ajaxFunction()' value="Login"></td>

    <tr>

    </form>

    <div id='ajaxDiv'>Your result will display here</div>

    </body>

    </html>

     

    my question is how to redirect to another page if the user and password is correct?

     

    Thanks for any help/ reply

     

     

     

     

  4. Hi,

    I have sample code like:

     

    $query = "INSERT INTO tbl_a1 VALUES ('','$item_id','$item_qty')";

    print "$query";

    mysql_query($query);

    header("location: item-list.php");

     

    the query is execute, comeback to item-list.php without interruption of print the query statement.

     

    Anyone know what is wrong with that situation/ code ?

    Thanks for any help

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