Ajax調用不返回期望值

Question

卡住爲什麼這不起作用，任何人有任何想法？它應該更新按鈕，我想這樣做，只是顯示的其他在結束所有的時間

function friendToggle(type,user,elem){ 
    var conf = confirm("Press Ok to confirm the '" +type+ "' action for user " + user); 
    if(conf != true){ 
     return false; 
    } 

    O(elem).innerHTML = 'please wait...'; 

    params = user 
    request = new ajaxRequest() 
    request.open("POST", "friendSystem.php", true) 
    request.setRequestHeader("Content-type", "application/x-www-form-urlencoded") 
    request.setRequestHeader("Content-length", params.length) 
    request.setRequestHeader("Connection", "close") 

    request.onreadystatechange = function(){ 
     if(this.readyState == 4){ 
      if(this.responseText == "friend_request_sent"){ 
       O(elem).innerHTML = 'OK Friend Request Sent'; 
      }else if(this.responseText == "unfriend_ok"){ 
       O(elem).innerHTML = '<button onclick="friendToggle(\'friend\',\'echo $view;\',\'friendBtn\')">Request As Friend</button>'; 
      }else { 
       alert(request.responseText); 
       O(elem).innerHTML = 'Try again later'; 
      } 
     } 
    } 
request.send("type="+type+"&user="+user); 

}

這是它調用的代碼：

if($_POST['type'] == "friend") 
{ 
    $friendCount = mysqli_fetch_row(querySQL("SELECT COUNT(*) FROM friends WHERE user='$user' AND accepted='1' OR friend='$user' AND accepted='1'")); 

    $blockCount1 = mysqli_fetch_row(querySQL("SELECT COUNT(*) FROM blockedusers WHERE blocker='$user' AND blockee='$view' LIMIT 1")); 

    $blockCount2 = mysqli_fetch_row(querySQL("SELECT COUNT(*) FROM blockedusers WHERE blocker='$view' AND blockee='$user' LIMIT 1")); 

    $rowCount1 = mysqli_fetch_row(querySQL("SELECT COUNT(*) FROM friends WHERE user='$view' AND friend='$user' AND accepted='1' LIMIT 1")); 

    $rowCount2 = mysqli_fetch_row(querySQL("SELECT COUNT(*) FROM friends WHERE user='$user' AND friend='$view' AND accepted='1' LIMIT 1")); 

    $rowCount3 = mysqli_fetch_row(querySQL("SELECT COUNT(*) FROM friends WHERE user='$view' AND friend='$user' AND accepted='0' LIMIT 1")); 

    $rowCount4 = mysqli_fetch_row(querySQL("SELECT COUNT(*) FROM friends WHERE user='$user' AND friend='$view' AND accepted='0' LIMIT 1")); 

    if($friendCount[0] > 99){ 
     echo"$user currently has the maximum number of friends and cannot accept any more"; 
    }elseif($blockCount1[0] > 0){ 
     echo "$user has you blocked, we cannot proceed."; 
    }elseif($blockCount2[0] > 0){ 
     echo"You must first unblock $user in order to friend them."; 
    }elseif($rowCount1[0] >0 || $rowCount2[0] > 0){ 
     echo"You are already friends with $user."; 
    }elseif($rowCount3[0] >0){ 
     echo"You have a pending friend request already sent to $user."; 
    }elseif($rowCount4[0] > 0){ 
     echo"$user has requested to friend with you first. Check your friend requests"; 
    }else{ 
     querySQL("INSERT INTO friends VALUES('$user','$view','0')"); 
     echo"friend_request_sent"; 
    } 

}  

它的意思回顯正確的錯誤信息，但沒有。也永遠不會發送好友請求發送消息

Answer 1

您的內容長度不正確。它被設置爲user的長度，但是您發送的數據比POST數據中的多。看起來Web服務器正在將您的數據縮減到錯誤的長度，然後無法分析格式錯誤的字符串。

讓你的POST數據到一個變量，這樣你可以得到的長度，而無需複製它：

var data = "type="+encodeURIComponent(type)+"&user="+encodeURIComponent(user); 
request.setRequestHeader("Content-length", data.length); 

然後當你調用發送：

request.send(data); 

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邊注：添加encodeURIComponent正確編碼這些值。