Question

如果ti留在数据库中,我想检查一些数据。例:以我的形式,有两种形式领域。 (1) 姓氏2 因此,我想检查一下,如果会议用户在数据库中已经掌握了这一词汇和名称,那么就必须显示错误的信息。这里只有几部神秘的法典。

<?php
$url        = mysql_real_escape_string(trim($_POST[ url ]));
$name       = mysql_real_escape_string(trim($_POST[ name ]));
$uname  = $_SESSION[ uname ];

$check_add  = mysql_query("SELECT * FROM add_fav WHERE url = $url  AND name=  $name  
AND uname =  $uname  ");

$num        =  mysql_num_rows($check_add); 

if($num > 0)
            $err[] = "Already exist";
?>

But It s doesn t check. May be it s my query problem.
Thanks in Advance:)

Answer 1

敬请和贬低你的文字,确保你实际上具有你的选择发言的价值。重复您的变量,以确保它们有数据...... 在你做选择发言之前,如果他们空洞,就可以确保你在接手之前即刻处理 d子关系。我sql_real_efall_string or 它赢得t work......

Answer 2

如果你与我的开诚布公,你是否会尝试?

你们是否在MySQL(如PpMyAdmin)直接尝试了这一询问?

你们应当使用PDO而不是我的sql_*,它会改善imp。

Here s an example, that, in code, should work (it then depend if your table is like what you show us):

<?php
$oPdo = new PDO( mysql:dbname=testdb;host=127.0.0.1 ,  login , password );

// here s your query :
$oStmt = $oPdo->prepare ( SELECT * FROM add_fav WHERE url = :url AND name= :name AND uname = :uname );

$oStmt->execute (array (
     url  => trim($_POST[ url ]),
     name  => trim($_POST[ name ]),
     uname  => $_SESSION[ uname ]
));

if (count($oStmt->fetchAll(PDO::FETCH_ASSOC)) > 0)
    $err[] = "Already exist";
?>

友情链接