php - PHP代码不会更新数据库

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“Notice: Undefined variable”, “Notice: Undefined index”, and “Notice: Undefined offset” using PHP
                                
                                    （28个答案）
                                
                        
                                3年前关闭。
            
                    
我正在做作业，但无法获取更新以使用数据库。
这是保存数据的表格。

<form name="edit" method="post" action="process/editRecord.php">

    <p class="indent">
      <label for="projectName">Edit Project Name</label>
      <input type="text" name="projectName" id="projectName" value="<?php echo $projectName; ?>">
    </p>

    <p class="indent">
      <label for="projectSoftware">Edit Project Software</label>
      <input type="text" name="projectSoftware" id="projectSoftware" value="<?php echo $projectSoftware; ?>" >
    </p>


    <p class="indent">
      <label for="projectDescription">Edit Project Description</label>
      <textarea name="projectDescription" id="projectDescription" cols="150" rows="10" ><?php echo $projectDescription; ?></textarea>
    </p>

    <p class="indent">
      <label for="projectImage">Edit Project Image</label>
      <input type="text" name="projectImage" id="projectImage" value="<?php echo $projectImage; ?>" >
    </p>

    <p class="indent">
      <label for="projectInformation">Edit Project Information</label>
      <textarea name="projectInformation" id="projectInformation" cols="400" rows="10" ><?php echo $projectInformation; ?></textarea>
    </p>

    <p>
      <input type="submit" name="button" id="button" value="Update">
    </p>

</form>

然后是过程。

<?php
ini_set('display_errors', 1);
require('../../includes/conn.inc.php');
require('../../includes/functions.inc.php');
// sanitize user variables
$sprojectName = safeString($_POST['projectName']);
$sprojectSoftware = safeString($_POST['projectSoftware']);
$sprojectDescription = safeString($_POST['projectDescription']);
$sprojectImage = safeString($_POST['projectImage']);
$sprojectInformation = safeString($_POST['projectInformation']);
$sprojectID = safeInt($_POST['projectID']);
// prepare SQL
$stmt = $mysqli->prepare("UPDATE projects SET projectName =?, projectSoftware =?, projectDescription=?, projectImage =?, projectInformation =? WHERE projectID = ?");
$stmt->bind_param('sssssi', $sprojectName, $sprojectSoftware, $sprojectDescription, $sprojectImage, $sprojectInformation, $sprojectID);
$stmt->execute();
$stmt->close();

header("Location: ../../php/projects.php");
// redirect browser
exit; // make sure no other code executed
?>

使用此工具时没有任何错误，它不会更新数据库，但会返回到相关项目页面。

最佳答案

这解决了问题


  $ _POST ['projectID']其格式不为– Dagon