Question

我没有收到错误，但它也没有返回行值。我该如何解决这个问题？

<?php 

    //DATABASE
    $dbConn     =   mysql_connect($host,$username,$password);
    mysql_select_db($database,$dbConn);

    $SQL =  mysql_query("SELECT N.*, C.CatID FROM News N INNER JOIN Categories C ON N.CatID = C.CatID WHERE N.Active = 1 ORDER BY DateEntered DESC");


    while ( $Result = mysql_fetch_array($SQL) or die(mysql_error())) {
    $CatID[] = $Result[CatID];
    $NewsName[] = $Result[NewsName];
    $NewsShortDesc[] = $Result[NewsShortDesc];
    }

//  mysql_free_result($Result);


?>
                <div class="toparticle">
                    <span class="section"><?=$CatID[0] ?> </span>
                    <span class="headline"><?=$NewsName[0] ?></span>
                    <p><?=$NewsShortDesc[0] ?></p>
                </div>

Answer 1

$Result[CatID];

需要

$Result['CatID'];

和

while ( $Result = mysql_fetch_array($SQL) or die(mysql_error())) {

需要

while ( $Result = mysql_fetch_array($SQL)) {

Answer 2

让我们正确地做到这一点如果由于临时存储以外的任何原因没有使用数组来循环显示，你可以直接将显示代码放入while循环。

<?php 
    //DATABASE
    $dbConn     =   mysql_connect($host,$username,$password);
    mysql_select_db($database,$dbConn);

    $SQL =  mysql_query("SELECT N.*, C.CatID FROM News N INNER JOIN Categories C ON N.CatID = C.CatID WHERE N.Active = 1 ORDER BY DateEntered DESC");

    while ( $Result = mysql_fetch_array($SQL) ) {
?>
                <div class="toparticle">
                    <span class="section"><?= $Result['CatID'] ?></span>
                    <span class="headline"><?= $Result['NewsName'] ?></span>
                    <p><?= $Result['NewsShortDesc'] ?></p>
                </div>
<?php
    }
    //mysql_free_result($Result);
?>

php查询无法弄清楚问题

2 个答案: