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I

Izu Great

@Izu Great
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  • need help with php parse error
    I Izu Great

    Thank your suggestions worked out.

    Web Development help php

  • need help with php parse error
    I Izu Great

    this is the problem.

    <?php
    $sql = "SELECT * FROM course_tb WHERE level LIKE '".$level."' AND semester LIKE '".$sem."'";
    if($result = $mysqli->query($sql)){
    if($result->num_rows > 0){
    while($rw = $result->fetch_array()){
    ?>

         <?
     }}}
     ?>
    

    whenver i try to loop through $result->fetch_array() it returns the Parse Error.

    Web Development help php

  • need help with php parse error
    I Izu Great

    Ok thanks I found the problem.

    Web Development help php

  • need help with php parse error
    I Izu Great

    i am working on a project and on a page i get this error. Parse error: parse error in C:\wamp\www\RMS\comresult.php on line 143 line 143 is the last line of my code. Please i need help. Here is my code:

    <?php
    include("conn.php");
    session_start();
    ?>
    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
    <html xmlns="http://www.w3.org/1999/xhtml">
    <head>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
    <title>Welcome :: Result Portal</title>
    <link href="style.css" rel="stylesheet" type="text/css" />
    <style type="text/css">
    body {
    background-color: #FFF;
    }
    </style>
    <!--[if IE]>
    <script src="ie/dist/html5shiv.js"></script>
    <script src="ie/dist/html5shiv-printshiv.js"></script>
    <![endif-->
    <script type="text/javascript">
    function bk(){
    location.assign(document.referrer);
    }
    var state = true;

    function getstate(){
    return state;
    }
    </script>
    </head>

    <body onload="print()">
    <?php
    if(isset($_GET['lev'],$_GET['sem'],$_GET['prog'],$_GET['sess'])){
    $level = $mysqli->real_escape_string(trim($_GET['lev']));
    $sem = $mysqli->real_escape_string(trim($_GET['sem']));
    $prog = $mysqli->real_escape_string(trim($_GET['prog']));
    $sess = $mysqli->real_escape_string(trim($_GET['sess']));

    $sql = "SELECT \* FROM course\_tb WHERE level LIKE '".$level."' AND semester LIKE '".$sem."'";
    if($result = $mysqli->query($sql)){
        if($result->num\_rows > 0){
    
        $sel = "SELECT DISTINCT regno FROM st\_result\_data WHERE level LIKE '".$level."' AND semester LIKE '".$sem."' AND programme LIKE '".$prog."'";
    
        if($res = $mysqli->query($sel)){
            if($res->num\_rows > 0){
    
                //$sel2 = "SELECT \* FROM st\_result\_data WHERE regno LIKE '".$row\['regno'\]."' AND level LIKE '".$level."' AND semester LIKE '".$sem."' AND programme LIKE '".$row\['programme'\].&
    
    Linux, Apache, MySQL, PHP javascript database help php html

  • need help with php parse error
    I Izu Great

    The braces are Matched correctly.

    Web Development help php

  • need help with php parse error
    I Izu Great

    Line 143 is the last line "". This is my code.

    <?php
    include("conn.php");
    session_start();
    ?>
    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
    <html xmlns="http://www.w3.org/1999/xhtml">
    <head>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
    <title>Welcome :: Result Portal</title>
    <link href="style.css" rel="stylesheet" type="text/css" />
    <style type="text/css">
    body {
    background-color: #FFF;
    }
    </style>
    <!--[if IE]>
    <script src="ie/dist/html5shiv.js"></script>
    <script src="ie/dist/html5shiv-printshiv.js"></script>
    <![endif-->
    <script type="text/javascript">
    function bk(){
    location.assign(document.referrer);
    }
    var state = true;

    function getstate(){
    return state;
    }
    </script>
    </head>

    <body onload="print()">
    <?php
    if(isset($_GET['lev'],$_GET['sem'],$_GET['prog'],$_GET['sess'])){
    $level = $mysqli->real_escape_string(trim($_GET['lev']));
    $sem = $mysqli->real_escape_string(trim($_GET['sem']));
    $prog = $mysqli->real_escape_string(trim($_GET['prog']));
    $sess = $mysqli->real_escape_string(trim($_GET['sess']));

    $sql = "SELECT \* FROM course\_tb WHERE level LIKE '".$level."' AND semester LIKE '".$sem."'";
    if($result = $mysqli->query($sql)){
        if($result->num\_rows > 0){
    
        $sel = "SELECT DISTINCT regno FROM st\_result\_data WHERE level LIKE '".$level."' AND semester LIKE '".$sem."' AND programme LIKE '".$prog."'";
    
        if($res = $mysqli->query($sel)){
            if($res->num\_rows > 0){
    
                //$sel2 = "SELECT \* FROM st\_result\_data WHERE regno LIKE '".$row\['regno'\]."' AND level LIKE '".$level."' AND semester LIKE '".$sem."' AND programme LIKE '".$row\['programme'\]."'";
            //if($res2 = $mysqli->query($sel2)){
                //if($res2->num\_rows > 0){
                ?>
    
    Web Development help php

  • need help with php parse error
    I Izu Great

    i am working on a project and on a page i get this error. Parse error: parse error in C:\wamp\www\RMS\comresult.php on line 143 line 143 is the last line of my code. Please i need help.

    Web Development help php

  • Help with mysql duplicate row
    I Izu Great

    Thanks. I'm using it on php

    Web Development database mysql help

  • Help with mysql duplicate row
    I Izu Great

    i have table in my database, which has theses fields id, regno, courseid, level, session, the regno can be inserted more than once as well as level and session, also courseid but with previous regno. i need to prevent duplicate rows with regno, courseid, level, session, on a row.

    Web Development database mysql help

  • Need to get the parameter from URl to build href.
    I Izu Great

    would want it in PHP? if yes:

    Web Development javascript html css sysadmin tools
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