duanfan1869 2014-05-08 19:06
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PHP找到了多少行

I have Google how to do this but I cant figure it out. So this is my code:

<?php
session_start();

Function PrintAll() {
    echo "<center>";
    echo "Filmpje Nummer: ".$_SESSION['hits'];
    echo "</center>";
    $dbnaam = "a2943462_Pages"; 

    $connection = mysql_connect($dbhost,$uname,$pass) or die ("niet gelukt om als $user te connecten met $host"); 
    $result = mysql_select_db($dbnaam,$connection);

    $sql = "SELECT * FROM PagesInfo WHERE ID=".$_SESSION['hits']; // sql select query 
    $result = mysql_query($sql, $connection); 

    while ($rij = mysql_fetch_array($result)) 
    { 
        echo "<center>";
        echo "<h3>" . $rij["Title"] . "</h3>";
        echo $rij["Video"] . "<br/>";
        echo "<p>" ."Posted By: " . $rij["Posted"] . "</p>";
        echo "</center>";
    } 
}

function ShowButtons() {
    echo "<html>
    <head><title>COUNTER</title>
    <link rel='stylesheet' type='text/css' href='style.css'/>
    </head>
    <body><center>
    <form  method='get'>
    <input type='submit' name='Previous' value='<< Previous'/>
    <input type='submit' name='Next' value='Next >>'/>
    </form></center>
    </body>
    </html>";
}

function DisPre() {
    echo "<html>
    <head><title>COUNTER</title>
    <link rel='stylesheet' type='text/css' href='style.css'/>
    </head>
    <body><center>
    <form  method='get'>
    <input type='submit' name='Next' value='Next >>'/>
    </form></center>
    </body>
    </html>";
}

if($_GET){
    $pageShow = $_SESSION['hits'];
    if(isset($_GET['Next'])){
        $_SESSION['hits']++;
        PrintAll();
    }elseif(isset($_GET['Previous'])){
        if ($_SESSION['hits'] < 1) {
            echo "Cannot go further Back";
            DisPre(); 
        }else{
            $_SESSION['hits']--;
            PrintAll();
        }
    }
}

if(isset($_SESSION['hits'])){
    if ($_SESSION['hits'] < 1){
        disPre();
    }else{
        ShowButtons();
    }
}else{
    $_SESSION['hits']=1;
    PrintAll();
}
?>

So how and where do i have to Add a code that does this:

If it doesn't find any rows in the Table PagesInfo then it will Echo this:

echo "<center><img src='Sourceher' height='390px' width='640px' /></center>";

Maybe someone can put the code inside my code ?? Any help is much appreciated.

  • 写回答

2条回答 默认 最新

  • doukefu1361 2014-05-08 19:13
    关注

    Change this portion:

    $sql = "SELECT * FROM PagesInfo WHERE ID=".$_SESSION['hits']; // sql select query 
    $result = mysql_query($sql, $connection); 
    
    while ($rij = mysql_fetch_array($result)) 
    { 
    echo "<center>";
    echo "<h3>" . $rij["Title"] . "</h3>";
    echo $rij["Video"] . "<br/>";
    echo "<p>" ."Posted By: " . $rij["Posted"] . "</p>";
    echo "</center>";
    }
    

    to:

    $sql = "SELECT * FROM PagesInfo WHERE ID=".$_SESSION['hits']; // sql select query 
    $result = mysql_query($sql, $connection); 
    if(mysql_num_rows($result) >0){
    while ($rij = mysql_fetch_array($result)) 
    { 
    echo "<center>";
    echo "<h3>" . $rij["Title"] . "</h3>";
    echo $rij["Video"] . "<br/>";
    echo "<p>" ."Posted By: " . $rij["Posted"] . "</p>";
    echo "</center>";
    }
    }
    else {
    echo "<center><img src='Sourceher' height='390px' width='640px' /></center>";
    
    }
    

    N.B: mysql_* is deprecated use PDO or mysqli_*.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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