沒有提示錯誤-無法從第二個表中提取數據

Question

2個表， users和useradvert （在sql中都具有良好的關系，只是不能從第二個表useradvert提取數據）。 與表useradvert （索引useradvert id ）相關的表User （主鍵id ）。 下面是摘錄。

//從表用戶和表用戶廣告中調用記錄，並加入

if(isset($_POST['username'])){
   $userName = $_POST['username']; 
   $query = "SELECT users.id, users.name, users.username, users.telno, useradvert.id, useradvert.name2, useradvert.color2, useradvert.hobby2, useradvert.radiobtn, useradvert.kupon, useradvert.image, useradvert.image2 ". "FROM users 
LEFT JOIN useradvert ON useradvert.id = users.id"." WHERE username= ?";   
   $stmt = $conn->prepare($query);
   $stmt->bind_param('s',$userName);
   $stmt->execute();
   $res = $stmt->get_result(); 
   $row = $res->fetch_array();
   $_SESSION['id'] = $row['id'];
   $_SESSION['name'] = $row['name'];
   $_SESSION['username'] = $row['username'];
   $_SESSION['telno'] = $row['telno'];
   $_SESSION['name2'] = $row['name2'];
   $_SESSION['color2'] = $row['color2'];
   $_SESSION['hobby2'] = $row['hobby2'];
   $_SESSION['radiobtn'] = $row['radiobtn'];
   $_SESSION['kupon'] = $row['kupon'];
   $_SESSION['image'] = $row['image'];
   $_SESSION['image2'] = $row['image2'];    
}
?>

繼續下面的摘錄...

</head>
<body>
<div id="apDiv3">
<p>&nbsp;</p>
<p>&nbsp;</p>
<p>&nbsp;</p>
<p><span class="TabbedPanelsContent">
  <?php
  //display record from table- users (parent table can display)
    echo $_SESSION['id']."<br/>";
    echo $_SESSION['name']."<br/>";
    echo $_SESSION['username']."<br/>";
    echo $_SESSION['telno']."<br/>";
?>

<?php
//display records from table -useradvert (child table cannot display)

while($row = $res->fetch_array()){
"<br/>";
"<br/>";
"<br/>";
    echo $_SESSION['id']."<br/>";
    echo $_SESSION['name2']."<br/>";
    echo $_SESSION['color2']."<br/>";
    echo $_SESSION['hobby2']."<br/>";
    echo $_SESSION['radiobtn']."<br/>";
    echo $_SESSION['kupon']."<br/>";
    echo $_SESSION['image']."<br/>";
    echo $_SESSION['image2']."<br/>";
}
?>

請幫忙

Answer 1

下面的答案（我正在分享此信息，以便世界上有相同問題的任何人都可以將此作為指導。）

if(isset($_POST['username'])){



    $userName = $_POST['username'];
    $query = "SELECT id, name, username, telno FROM users WHERE username = ?";
    $stmt = $conn->prepare($query);
    $stmt->bind_param('s', $userName);
    $stmt->execute();
    $res = $stmt->get_result(); 
    $row = $res->fetch_array();
    $_SESSION['id'] = $row['id'];
    $_SESSION['name'] = $row['name'];
    $_SESSION['username'] = $row['username'];
    $_SESSION['telno'] = $row['telno'];



    $query = "SELECT useradvert.name2, useradvert.color2, useradvert.hobby2,useradvert.radiobtn, useradvert.kupon, useradvert.image, useradvert.image2 FROM users INNER JOIN useradvert ON users.id=useradvert.id ";

    $stmt = $conn->prepare($query); 
    $stmt->execute();
    $res = $stmt->get_result(); 
    $row2 = $res->fetch_array();

  // $_SESSION['name'] = $row2['name'];
    $_SESSION['name2'] = $row2['name2'];
    $_SESSION['color2'] = $row2['color2'];
    $_SESSION['hobby2'] = $row2['hobby2'];
    $_SESSION['radiobtn'] = $row2['radiobtn'];
    $_SESSION['kupon'] = $row2['kupon'];
    $_SESSION['image'] = $row2['image'];
    $_SESSION['image2'] = $row2['image2'];

}
?>

在同一頁面上提取以下內容。

  <?php
  //display record from table- users (parent table)
    echo $_SESSION['id']."<br/>";
    echo $_SESSION['name']."<br/>";
    echo $_SESSION['username']."<br/>";
    echo $_SESSION['telno']."<br/>";
?>

在同一頁面上提取以下內容。

 <?php
  //display record from table- useradveret -(child table)
  while($row = $res->fetch_array()){
   // echo $_SESSION['name']."<br/>";
    echo $_SESSION['name2']."<br/>";
    echo $_SESSION['color2']."<br/>";
    echo $_SESSION['hobby2']."<br/>";
    echo $_SESSION['radiobtn']."<br/>";
    echo $_SESSION['kupon']."<br/>";
    echo $_SESSION['image']."<br/>";
    echo $_SESSION['image2']."<br/>";}

?>

沒有提示錯誤-無法從第二個表中提取數據

問題描述

1 個解決方案

解決方案1
0 2016-01-05 08:05:25

沒有提示錯誤-無法從第二個表中提取數據

問題描述

1 個解決方案

解決方案1 0 2016-01-05 08:05:25

解決方案1
0 2016-01-05 08:05:25