php显示未显示的记录_Php - Fatal编程技术网

php显示未显示的记录

php

php显示未显示的记录,php,Php,我试图用mysqli用php显示记录。这是我的php显示记录代码，它不会给出任何错误，但也不会显示任何记录。插入操作正在工作，但显示不工作此处缺少什么 <!DOCTYPE html> <html> <body> <form method="POST" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>"> movie_name<input ty

我试图用mysqli用php显示记录。这是我的php显示记录代码，它不会给出任何错误，但也不会显示任何记录。插入操作正在工作，但显示不工作此处缺少什么

    <!DOCTYPE html>
<html>
<body>

<form method="POST" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>">

   movie_name<input type="text" name="movie_name"><br>
   movie_embedded_id<input type="text" name="movie_embedded_id"><br>
   rating_no<input type="text" name="rating_no"><br>
   movie_description<input type="text" name="movie_description"><br>
   <input type="submit" name="submit" value="Submit"> 

</form>

<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname="mydb";

// Create connection
$conn = new mysqli($servername, $username, $password,$dbname);

// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 
echo "Connected successfully";

        if(isset($_POST['movie_name']) && $_POST['movie_name'] != "")
            {   
                $sql = "INSERT INTO movies (movie_name, movie_embedded_id, rating_no, movie_description)VALUES ('".$_POST['movie_name']."', '".$_POST['movie_embedded_id']."', '".$_POST['rating_no']."','".$_POST['movie_description']."')";


            if ($conn->query($sql) === TRUE) {
            echo "New record created successfully";
            } else {
            echo "Error: " . $sql . "<br>" . $conn->error;
            }

        }

$conn->close();


?>


<?php

      if(isset($_POST['movie_name']) && $_POST['movie_name'] != "")
            {
$sql = "select * from movies where movie_id = ".$_GET["id"]."";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
     echo "<table><tr><th>ID</th><th>movie_name</th><th>movie_embedded_id</th><th>rating_no</th><th>movie_description</th></tr>";
     // output data of each row
     while($row = $result->fetch_assoc()) {
         echo "<tr><td>" . $row["id"]. "</td><td>" . $row["movie_name"]. " </td><td>" . $row["movie_embedded_id"]. " </td><td>" . $row["rating_no"]. "</td><td> " . $row["rating_no"]. "</td></tr>";
     }
     echo "</table>";
} else {
     echo "0 results";
}

    }


  ?>


</body>
</html>


让我们试试这个
它会起作用的
 <?php

        if(isset($_POST['movie_name']) && $_POST['movie_name'] != "")
            {
$sql = "select * from movies where movie_id = ".$_GET['id'];
$result = $conn->query($sql);

if ($result->num_rows > 0) {
     echo "<table><tr><th>ID</th><th>movie_name</th><th>movie_embedded_id</th><th>rating_no</th><th>movie_description</th></tr>";
     // output data of each row
     while($row = $result->fetch_assoc()) {
         echo "<tr><td>" . $row["id"]. "</td><td>" . $row["movie_name"]. " </td><td>" . $row["movie_embedded_id"]. " </td><td>" . $row["rating_no"]. "</td><td> " . $row["rating_no"]. "</td></tr>";
     }
     echo "</table>";
} else {
     echo "0 results";
}

    }

非常感谢大家的友好回复。这里我已经解决了我的问题。只有当我添加id列时出现问题，额外的空列被添加，不知道为什么。请查看此代码以获得更安全和优化
    <div class="container">
<form class="form-horizontal" method="POST" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>">
  <div class="form-group">
    <label class="control-label col-sm-2" for="movie_name">movie_name:</label>
    <div class="col-sm-10">
      <input type="text" class="form-control" id="movie_name" name="movie_name" placeholder="Enter movie name">
    </div>
  </div>
  <div class="form-group">
    <label class="control-label col-sm-2" for="movie_embedded_id">movie_embedded_id:</label>
    <div class="col-sm-10"> 
      <input type="text" class="form-control" id="movie_embedded_id" name="movie_embedded_id" placeholder="Enter embeded id">
    </div>
  </div>
  <div class="form-group">
    <label class="control-label col-sm-2" for="rating_no">rating_no:</label>
    <div class="col-sm-10"> 
      <input type="text" class="form-control" id="rating_no" name="rating_no" placeholder="Enter rating_no">
    </div>
  </div>
    <div class="form-group">
    <label class="control-label col-sm-2" for="movie_description">rating_no:</label>
    <div class="col-sm-10"> 
      <input type="text" class="form-control" id="movie_description" name="movie_description" placeholder="Enter movie_description">
    </div>
  </div>
  <div class="form-group"> 
    <div class="col-sm-offset-2 col-sm-10">
      <button type="submit" name="submit" class="btn btn-default" >submit</button>
    </div>
  </div>
</form>
</div>
  <table class="table table-bordered">
    <thead>
      <tr>
<!--       <th>ID.<th> -->
       <th>movie_name</th>
       <th>movie_embedded_id</th>
       <th>rating_no</th>
       <th>movie_description</th>
       <th>Update</th>
      </tr>
    </thead>
        <tbody>
    <?php  
    include "db.php";
    include "movieinsert.php";

      $result ="select * from movies";
      $get=mysqli_query($conn, $result);

    while ($data = mysqli_fetch_array($get))
        {
    ?>
          <tr>
    <!--      <td><?php //echo "$data[0]"; ?><td> -->
          <td><?php echo "$data[1]"; ?></td>
          <td><?php echo "$data[2]"; ?></td>
          <td><?php echo "$data[3]"; ?></td>
          <td><?php echo "$data[4]"; ?></td>
          <td>
          <a href='mvieupdate.php?id=$data[0]' class="btn btn-info" role="button">Update</a>
          </td>
          </tr>
    <?php 
         }
     ?>
  <tbody>
   </table>


你检查过你的源代码了吗？可能他创建了表标记，但是$row数据只是空的。如果您调试代码，意味着$\u POST['movie\u name']设置了吗？还为$\u POST['movie\u name']添加了else条件，并回显一个正常值msg@AmarGhodke只需尝试替换此查询“select*from movies where movie_id=”。$\u GET['id']。“”；对于此“从电影中选择*，其中电影id=“.$\u获取[“id”]”。@Bunker Boy抱歉，这不起作用，请允许我在此处提供完整代码，您可能会发现实际问题。请尝试此$sql=“从电影中选择*，其中电影id=“.$\u获取[“id””；
    <div class="container">
<form class="form-horizontal" method="POST" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>">
  <div class="form-group">
    <label class="control-label col-sm-2" for="movie_name">movie_name:</label>
    <div class="col-sm-10">
      <input type="text" class="form-control" id="movie_name" name="movie_name" placeholder="Enter movie name">
    </div>
  </div>
  <div class="form-group">
    <label class="control-label col-sm-2" for="movie_embedded_id">movie_embedded_id:</label>
    <div class="col-sm-10"> 
      <input type="text" class="form-control" id="movie_embedded_id" name="movie_embedded_id" placeholder="Enter embeded id">
    </div>
  </div>
  <div class="form-group">
    <label class="control-label col-sm-2" for="rating_no">rating_no:</label>
    <div class="col-sm-10"> 
      <input type="text" class="form-control" id="rating_no" name="rating_no" placeholder="Enter rating_no">
    </div>
  </div>
    <div class="form-group">
    <label class="control-label col-sm-2" for="movie_description">rating_no:</label>
    <div class="col-sm-10"> 
      <input type="text" class="form-control" id="movie_description" name="movie_description" placeholder="Enter movie_description">
    </div>
  </div>
  <div class="form-group"> 
    <div class="col-sm-offset-2 col-sm-10">
      <button type="submit" name="submit" class="btn btn-default" >submit</button>
    </div>
  </div>
</form>
</div>
  <table class="table table-bordered">
    <thead>
      <tr>
<!--       <th>ID.<th> -->
       <th>movie_name</th>
       <th>movie_embedded_id</th>
       <th>rating_no</th>
       <th>movie_description</th>
       <th>Update</th>
      </tr>
    </thead>
        <tbody>
    <?php  
    include "db.php";
    include "movieinsert.php";

      $result ="select * from movies";
      $get=mysqli_query($conn, $result);

    while ($data = mysqli_fetch_array($get))
        {
    ?>
          <tr>
    <!--      <td><?php //echo "$data[0]"; ?><td> -->
          <td><?php echo "$data[1]"; ?></td>
          <td><?php echo "$data[2]"; ?></td>
          <td><?php echo "$data[3]"; ?></td>
          <td><?php echo "$data[4]"; ?></td>
          <td>
          <a href='mvieupdate.php?id=$data[0]' class="btn btn-info" role="button">Update</a>
          </td>
          </tr>
    <?php 
         }
     ?>
  <tbody>
   </table>