更新查询不起作用
更新查询不起作用
提问于 2019-03-22 04:07:09
我正在尝试开发一个小型的“待办事项列表”应用程序。应用程序的数据存储在数据库中，它需要执行所有CRUD操作。现在，Select、Insert和Delete都可以正常工作。但是我被困在更新中了。index.php页面如下所示：
<?php
session_start();
require_once 'connect.php';
if (isset($_POST['DeleteTask'])) {
$sqldelete = "DELETE FROM Tasks WHERE dbTaskID = :bvTaskID";
$stmtdelete = $db->prepare($sqldelete);
$stmtdelete->bindValue(':bvTaskID', $_POST['taskID']);
$stmtdelete->execute();
echo "<div>Task successfully deleted</div>";
}
if (isset($_POST['theSubmit'])){
echo '<p>New task added</p>';
$formfield['ffTaskName'] = trim($_POST['taskName']);
$formfield['ffTaskDue'] = trim($_POST['taskDue']);
if(empty($formfield['ffTaskName'])){$errormsg .= "<p>Task field is empty.</p>";}
if(empty($formfield['ffTaskDue'])){$errormsg .= "<p>Deadline field is empty.</p>";}
if ($errormsg != "") {
echo "<div class='error'><p>Please fill out all fields before submitting.</p>";
echo $errormsg;
echo "</div>";
} else {
try {
$sqlinsert = 'INSERT INTO Tasks (dbTaskName, dbTaskDue, dbTaskDone)
VALUES (:bvTaskName, :bvTaskDue, :bvTaskDone)';
$stmtinsert = $db->prepare($sqlinsert);
$stmtinsert->bindValue(':bvTaskName', $formfield['ffTaskName']);
$stmtinsert->bindValue(':bvTaskDue', $formfield['ffTaskDue']);
$stmtinsert->bindValue(':bvTaskDone', 0);
$stmtinsert->execute();
echo "<div><p>There are no errors. Thank you.</p></div>";
} catch(PDOException $e){
echo 'ERROR!!!' .$e->getMessage();
exit();
}
}
}
$sqlselect = "SELECT * from Tasks";
$result = $db->prepare($sqlselect);
$result->execute();
?>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>To Do Application</title>
</head>
<body>
<h1><u>To-Do List</u></h1>
<table border>
<tr>
<th>Task</th>
<th>Deadline</th>
<th>Status</th>
<th>Complete</th>
<th>Edit</th>
<th>Delete</th>
</tr>
<?php
while ($row = $result->fetch()) {
if ($row['dbTaskDone'] == 0) {
$status = "Unfinished";
} else {
$status = "Finished";
}
echo '<tr><td>' . $row['dbTaskName']
. '</td><td>' . $row['dbTaskDue']
. '</td><td>' . $status;
/*if ($status == "Unfinished"){
echo '</td><td>';
echo '<form action="'. $_SERVER['PHP_SELF'] . '" method="post">';
echo '<input type="hidden" name="taskID" value"' . $row['dbTaskID'] . '">';
echo '<input type="submit" name="CompleteTask" value="Complete Task">';
echo '</form>';
}*/
echo '</td><td>';
echo '<form action="updateTask.php" method="post">';
echo '<input type="hidden" name="taskID" value="' . $row['dbTaskID'] . '">';
echo '<input type="submit" name="EditTask" id="EditTask" value="Edit Task">';
echo '</form></td><td>';
echo '<form action="'. $_SERVER['PHP_SELF'] . '" method="post">';
echo '<input type="hidden" name="taskID" value="' . $row['dbTaskID'] . '">';
echo '<input type="submit" name="DeleteTask" value="Delete Task">';
echo '</td></tr>';
}
?>
</table>
<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>" name="toDoForm">
<fieldset><legend>New Task</legend>
<table>
<tr>
<th>Task</th>
<td><input type="text" name="taskName" id="taskName"
value="<?php echo $formfield['ffTaskName']; ?>"></td>
</tr><tr>
<th>Deadline</th>
<td><input type="text" name="taskDue" id="taskDue"
value="<?php echo $formfield['ffTaskDue']; ?>"></td>
</tr>
</table>
<input type="submit" name = "theSubmit" value="Add Task">
</fieldset>
</form>
</body>
</html>每条记录显示一个“编辑”按钮，该按钮从"Tasks“表中抓取PK并将其发送到updateTask.php页面：
<?php
require_once 'connect.php';
$errormsg = "";
if (isset($_POST['EditTask']) ) {
$formfield['ffTaskID'] = $_POST['taskID'];
$sqlselect = "SELECT * FROM Tasks WHERE dbTaskId = :bvTaskID";
$result = $db->prepare($sqlselect);
$result->bindValue(':bvTaskID', $formfield['ffTaskID']);
$result->execute();
$row = $result->fetch();
if( isset($_POST['theEdit']) )
{
$formfield['ffTaskID'] = $_POST['taskID'];
$formfield['ffTaskName'] = trim($_POST['taskName']);
$formfield['ffTaskDue'] = trim($_POST['taskDue']);
if(empty($formfield['ffTaskName'])){$errormsg .= "<p>Task field is empty.</p>";}
if(empty($formfield['ffTaskDue'])){$errormsg .= "<p>Deadline field is empty.</p>";}
if ($errormsg != "") {
echo "<div class='error'><p>Please fill out all fields before submitting.</p>";
echo $errormsg;
echo "</div>";
} else {
try
{
$sqlUpdate = "UPDATE Tasks SET dbTaskName = :bvTaskName,
dbTaskDue = :bvTaskDue
WHERE dbTaskID = :bvTaskID";
$stmtUpdate = $db->prepare($sqlUpdate);
$stmtUpdate->bindvalue(':bvTaskName', $formfield['ffTaskName']);
$stmtUpdate->bindvalue(':bvTaskDue', $formfield['ffTaskDue']);
$stmtUpdate->bindvalue(':bvTaskID', $formfield['ffTaskID']);
$stmtUpdate->execute();
}
catch(PDOException $e)
{
echo 'ERROR!!!' .$e->getMessage();
exit();
}
}
}
}
?>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>To Do Application</title>
</head>
<body>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post" name="editForm">
<fieldset><legend>Edit Task</legend>
<table>
<tr>
<th>Task</th>
<td><input type="text" name="taskName" id="taskName"
value="<?php echo $row['dbTaskName'];?>" ></td>
</tr><tr>
<th>Deadline</th>
<td><input type="text" name="taskDue" id="taskDue"
value="<?php echo $row['dbTaskDue']; ?>"></td>
</tr>
<tr>
<th>Submit Changes</th>
<input type="hidden" name="taskID" value="<?php echo $_formfield['ffTaskID']; ?>">
<td><input type="submit" name="theEdit" value="Submit Changes">
</table>
</fieldset>
</form>
</body>
</html>Name和Deadline字段根据从上一页传递的pk值相应地填充。但是，每当我按下“提交更改”按钮时，更新似乎不会执行。页面只是刷新，我看到数据库中的表数据保持不变。
回答 1
Stack Overflow用户
发布于 2019-03-22 05:40:17
解决了问题！
我发现了几个问题。
1.)在updateTask.php中，我将第二个if语句嵌套在第一个语句中。因此，它在页面加载时运行更新查询，而不更改数据。因此，'theEdit‘按钮没有做任何事情，因为它需要前一个if语句的条件才能运行。
2.)updateTask.php表单底部的表单字段'ffTaskID‘键入错误，该表单字段将在按下'theEdit’按钮时传递。
$_formfield
..should已经..
$formfield
此时，update查询功能正常。
3.)“编辑”按钮的问题已修复。尽管老实说我不能确定它是如何修复的。这可能与问题的第一部分有关。因此，当这个问题得到解决时，这个问题也就解决了。
无论哪种方式，一切似乎都在发挥应有的作用。再次感谢所有发表评论和提供帮助的人。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/55288507
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