duanbichou4942 2014-05-07 22:25
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已采纳

MySQL Query不起作用PHP [关闭]

$result = mysql_query("
    select CashID,something,something1,something2 
    from MsUser mu, MsCash mc 
    where mu.UserID = mc.UserID and (something2 = 1 or something2 = 2)");

while(($row = mysql_fetch_array($result) !== false)){
    This doesn't works
}

CashID is in MsCash, UserID is in MsUser and MsCash. Why it doesn't work on PHP? It works when try to take only one table, but when its get joined, mysql_fetch_array return me an error.

$resultLog = mysql_query("
    select UserID from MsUser where username=$username and password=$password 
    ");

while(($row1 = mysql_fetch_array($resultLog) !== false)){
    //This works
}

Warning: mysql_fetch_array() expects parameter 1 to be resource, null given ...

I have tried this in a blank page

<?
include "connect.php";

$result = mysql_query("
    select CashID,something,something1,something2 
    from MsUser mu, MsCash mc 
    where mu.UserID = mc.UserID and (something2 = 1 or something2 = 2)");
if($result === false){
    echo 'Query error: '.mysql_error();
}else{
    echo "works";
}
?>

but it give me nothing, also the echo doesn't show too.

Thanks.

  • 写回答

1条回答 默认 最新

  • drmy1050 2014-05-07 22:27
    关注
    SELECT CashID,something,something1,something2
    FROM MsUser mu
    LEFT JOIN MsCash mc ON mu.UserID = mc.UserID
    WHERE something2 = 1 or something2 = 2
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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