Question

我不明白为什么我无法从我的数据库中获取任何数据。它与$ userRow一起工作正常，但是当我使用$ SQL时它将无法正常工作。我很抱歉，如果它很容易修复，但我找不到我的错误

 // check if the form has been submitted and display the results
    if (isset($_POST['studentnum'])) {


      $conn = mysqli_connect($DBhost, $DBuser, $DBpass, $DBname);
      if (!$conn) {
        die('Could not connect: ' . mysqli_connect_error());
      }

      // escape the post data to prevent injection attacks
      $studentnum = mysqli_real_escape_string($conn, $_POST['studentnum']);

      $sql = "SELECT * FROM `afventer` WHERE `user_id` LIKE '%$studentnum%'"; 
      $result=mysqli_query($conn, $sql);

      // check if the query returned a result
      if (!$result) {
          echo 'Der var ikke nogen resultater på din søgning';
      } else {
        // result to output the table
        echo "Du har aftalt at køre"; 
        while($row = mysqli_fetch_array($result, MYSQLI_ASSOC))
        {  
          echo "&nbsp;"; 
          echo $row['uname'];
          echo "&nbsp;";
          echo "hjem";
        }
      }

当我回显SQL时它显示SELECT * FROM afventer WHERE user_id LIKE '%3%'，所以当我在数据库中的SQL中运行该字符串时，它会找到该用户，但它仍然不会显示在网站上

它主要是数据库正在做它想做的事情，但是我要求显示在页面上的数据没有显示

Answer 1

我通过将代码重写到此

来修复它

/ check if the form has been submitted and display the results
    if (isset($_POST['studentnum'])) {


      $conn = mysqli_connect($DBhost, $DBuser, $DBpass, $DBname);
      if (!$conn) {
        die('Could not connect: ' . mysqli_connect_error());
      }

      // escape the post data to prevent injection attacks
      $studentnum = mysqli_real_escape_string($conn, $_POST['studentnum']);

      $sql = "SELECT * FROM `afventer` WHERE `user_id` LIKE '%$studentnum%'"; 
      $result=mysqli_query($conn, $sql);

      // check if the query returned a result
      if (!$result) {
          echo 'Der var ikke nogen resultater på din søgning';
      } else {
        // result to output the table
        echo "Du har aftalt at køre"; 
        while($row = mysqli_fetch_array($result, MYSQLI_ASSOC))
        {  
          echo "&nbsp;"; 
          echo $row['username'];
          echo "&nbsp;";
          echo "hjem";
        }
        $conn = mysqli_connect($DBhost, $DBuser, $DBpass, $DBname);
  if (!$conn) {
    die('Could not connect: ' . mysqli_connect_error());
  }
  // escape the post data to prevent injection attacks
  $studentnum = mysqli_real_escape_string($conn, $_POST['studentnum']);

  $sql = "SELECT * FROM `afventer` WHERE `user_id` LIKE '%$studentnum%'"; 
  $result=mysqli_query($conn, $sql);
  $row = mysqli_fetch_array($result, MYSQLI_ASSOC);

 $parent = $userRow['user_id'];
 $parent_name = $userRow ['uname'];
 $student_name = $row ['username'];
 $mellem = $row['adresse'];
 $slut = $userRow['adresse'];

 $parent = $DBcon->real_escape_string($parent);
 $parent_name = $DBcon->real_escape_string($parent_name);
 $student_name = $DBcon->real_escape_string($student_name);
 $mellem =  $DBcon->real_escape_string($mellem);
 $slut = $DBcon->real_escape_string($slut);

 if ($count==0) {

  $sql = "INSERT INTO aftalt(student_id, student_name, parent_id, parent_name, mellem_adresse, slut_adresse) VALUES('$studentnum', '$student_name', '$parent', '$parent_name', '$mellem', '$slut')";
    $sql = "DELETE FROM afventer WHERE username = '$student_name'";
    echo "$sql";
  if ($DBcon->query($sql)) {
   $msg = "";
} else {
    echo "Error deleting record: " . mysqli_error($conn);
}
 }
      }

我无法从我的数据库中获取数据

1 个答案: