没有提示错误-无法从第二个表中提取数据

Question

2 tables, users and useradvert (both have relation in sql working fine, just cannot pull data from second table useradvert ). 2个表， users和useradvert （在sql中都具有良好的关系，只是不能从第二个表useradvert提取数据）。 table User (Primary key- id ) related to table useradvert (index key- id ). 与表useradvert （索引useradvert id ）相关的表User （主键id ）。 below is extract. 下面是摘录。

//calling for record from table-users and table-useradvert, joined //从表用户和表用户广告中调用记录，并加入

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'];    
}
?>

continue extract below... 继续下面的摘录...

</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/>";
}
?>

Please help 请帮忙

Answer 1

answers below (i am sharing this so anyone who has the same problem in the world can refer this as a guidance.) 下面的答案（我正在分享此信息，以便世界上有相同问题的任何人都可以将此作为指导。）

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'];

}
?>

extract below on same page.. 在同一页面上提取以下内容。

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

extract below on same page.. 在同一页面上提取以下内容。

 <?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/>";}

?>

没有提示错误-无法从第二个表中提取数据

问题描述

1 个解决方案

解决方案1
0 2016-01-05 08:05:25

没有提示错误-无法从第二个表中提取数据

问题描述

1 个解决方案

解决方案1 0 2016-01-05 08:05:25

解决方案1
0 2016-01-05 08:05:25