警告：mysqli_fetch_row（）预计参数1被mysqli_result

Question

我很抱歉，我问的问题，我与PHP和MySQL相当新的，我是相当新的理解调试。我创建该加法构件/朋友文件并且处理文件有错误

Warning: mysqli_fetch_row() expects parameter 1 to be mysqli_result, boolean given in line 57 

这是$countArr = mysqli_fetch_row($countResult2);

我也收到此错误

Warning: Cannot modify header information - headers already sent by 

这是标头：位置部分，我搜索如何解决这个问题，但我不明白 如果这是解决它的正确方法，如何实现ob_start();代码。

任何人都知道它为什么是这样呢？如果是这样，你能否好好解释一下。谢谢你这么多

下面

是代码，称为memberaddprocess.php

<?php 
////////////////////////////////////// 
    session_start(); 
    require_once('sqlconnect.inc.php'); 
///////////////////////////////////// 

    $memberHid1 = $_POST['memberHid1']; 
    $new_session = $_POST['memberHid']; 
//connecting to the database 
$conn = @mysqli_connect($host, 
     $user, 
     $pswd, 
     $dbnm); 

    if (!$conn) { 
     echo "<p>Database connection failure</p>"; 
    } else { 

    @mysqli_select_db($conn, $dbnm) 
     or die ("Database not available"); 
    } 

    $query = "SELECT member_id 
       FROM team 
       WHERE member_name = '$memberHid1'"; 

    $queryResult = @mysqli_query($conn, $query) 
       or die ("<p>Unable to execute query.</p>". "<p>Error code:" . mysqli_errno($conn) .":" . mysqli_error($conn)); 

    $fetchArr = mysqli_fetch_row($queryResult); 

    $memberHid3 = $fetchArr[0];    
    $memberHid = $_SESSION['membername'] = $new_session; 

    $query2 = "SELECT member_id 
       FROM team 
       WHERE member_name = '$memberHid'"; 

    $queryResult2 = @mysqli_query($conn,$query2) 
     or die ("<p>Unable to execute query.</p>". "<p>Error code" . mysqli_errno($conn) .":" . mysqli_error($conn)); 

    $fetchArr2 = mysqli_fetch_row($queryResult2); 
    $memberHid4 = $fetchArr2[0]; 

    $query3 = "INSERT INTO myteam 
       VALUES($memberHid4, $memberHid3)"; 

    $queryResult3 = @mysqli_query($conn,$query3) 
     or die ("<p>Unable to execute query.</p>". "<p>Error code" . mysqli_errno($conn) .":" . mysqli_error($conn))."</p>"; 
       echo "<p>$memberHid1"." "." Successfully added</p>"; 

     $queryCount = "SELECT COUNT(*) 
         FROM team"; 
     $countResult = @mysqli_query($conn,$queryCount); 
     $fetchCountArr = mysqli_fetch_row($countResult);  
     for($n=0;$n<$fetchCountArr[0];$n++) 
     { 
      $pst = $n+1; 
      $query4 = "SELECT member_id2 
         FROM myteam 
         WHERE friend_id1 = '$pst'"; 

      $countResult2 = @mysqli_query($conn,$query4); 
      $countArr = mysqli_fetch_row($countResult2); 
      $a=0; 
      while($countArr) 
      { 
       $a++; 
       $countArr = mysqli_fetch_row($countResult2); //echo "<p>$a</p>"; 
      } 

      $query4 = "UPDATE team 
         SET num_of_members= '$a' 
         WHERE member_id = '$pst'"; 

      $countResult2 = @mysqli_query($conn,$query4); 

     } 

    $querySelect = "SELECT member_id2 
        FROM myteam 
        WHERE member_id1 = '$memberHid4'"; 

    $querySelectResult = @mysqli_query($conn, $querySelect); 
    $fetchArr = mysqli_fetch_row($querySelectResult); 

    while($fetchArr) 
    { 
     foreach($fetchArr as $value) 

     { 
      //echo $value; 
      $querySelect2 = "SELECT member_name 
          FROM friends where friend_id='$value'"; 
      $querySelectResult2 = @mysqli_query($conn, $querySelectResult2); 
      $fetchArr2 = mysqli_fetch_row($querySelectResult2); 
      foreach($fetchArr2 as $value) 
      { 
       //echo $value; 
      }  
     } 
     $fetchArr = mysqli_fetch_row($querySelectResult); 
    } 

    header('Location: memberadd.php'); //redirecting to home page 
    exit(); 

?> 

Answer 1

我认为这将有助于你

<?php 
      $query4 = "SELECT member_id2 
        FROM myteam 
        WHERE friend_id1 = '$pst'"; 

     $countResult2 = @mysql_query($query4,$conn); 
     $countArr = mysql_fetch_array($countResult2); 
     echo $countArr['member_id2']; 
    ?>