Question

我试图更新与JQuery/JSON的记录,但发现这一错误:

发生错误:

[目标]

副市长

辛塔克斯·埃多尔:

My JS:

jQuery( #commentForm ).live( submit , function (event) {
event.preventDefault()

    
jQuery.ajax(edit_url, {
    data: jQuery(this).serialize(),
    dataType:  json ,
    type:  POST ,
    success: function (data) {
        if (data.error ===  OK ) {
            alert( ok c good )
        } else {
            alert( hi  + data.error)
        }
    },
    error: function(x,y,z){
        alert( 发生错误:
  + x +  
  + y +  
  + z);
    }
})
    
return false;
})

www.un.org/spanish/ecosoc And my php:

$ret = array(
     error              =>   OK ,
);
$update =
    "UPDATE crm_set_users SET ".
        "crm_set_users_civilite =  ".mysql_real_escape_string($crm_set_users_civilite)." ,".
        "crm_set_users_nom =  ".mysql_real_escape_string($crm_set_users_nom)." ,".
        "crm_set_users_prenom =  ".mysql_real_escape_string($crm_set_users_prenom)." ,".
        "crm_set_users_email =  ".mysql_real_escape_string($crm_set_users_email)." , ".
        "crm_set_users_telephone =  ".mysql_real_escape_string($crm_set_users_telephone)." , ".
        "crm_set_users_portable =  ".mysql_real_escape_string($crm_set_users_portable)." ";
        
if($crm_set_users_photo != ""){
    $update .=", crm_set_users_photo =  ".mysql_real_escape_string($crm_set_users_photo)." ";
}   
    
$update .=
    "WHERE ".
        "crm_set_users_id =  ".mysql_real_escape_string($user_id)." ";




echo json_encode($ret);
exit;

如果我是:

$ret = array(
     error              =>   OK ,
);
echo json_encode($ret);
exit;

之后,它......

请你们帮助!

Answer 1

或许,你的PHP投下了一些错误或警告,使归还的文件不能成为有效的JSON。参看以下表格:http://getfirebug.com/“rel=“nofollow”>FireBug,了解实际结果由您的PHP书写。

Following the PHP docu for mysql_real_escape_string (link)it will throw an Error, if there is no active MySQL connection available. Maybe that s your problem.

Answer 2

问题必须是,更新记录的工作失败了,因此json没有产出。

I think you need to add a space before the WHERE

$update .=
    " WHERE ".

在护堤后给它一些房间

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