我正在尝试创建一个表单,该表单将根据用户从选项下拉列表中选择的属性删除表中的一行。出于某种原因,第一个选项 (attemptid) 是一个 int,可以工作,但其他三个(它们是 varchar)不能。我为调试脚本而设置的错误处理正在返回1
or true
,但有问题的行没有被删除。
帮助!我已经尝试了一切,但只是在学习 PHP,所以想象一下我错过了一些非常简单的东西。
require_once("settings.php");
$conn = @mysqli_connect($host, $user, $pass, $db);
if ($conn) {
?>
<form method="post" action="delete_attempts.php" name="delete_attempts" id="delete_attempts" >
<label for="deleteby">
<p>Select an option to delete results by:</p>
</label>
<select name="deleteby">
<option value="attemptid">Attempt ID</option>
<option value="firstname">First Name</option>
<option value="lastname">Last Name</option>
<option value="studentid">Student ID</option>
</select>
<p></p>
<input type="text" name="delvalue" placeholder="Value">
<div>
<input type="submit" value="Delete Record" id="submit" />
</div>
</form>
<?php
if (isset($_POST["delvalue"])) {
// get value from form
$delValue = trim($_POST["delvalue"]);
echo $delValue;
// queries to delete record
$queryAttemptId = "DELETE FROM quizattempts WHERE attemptid = '$delValue'";
$queryFirstName = "DELETE FROM quizattempts WHERE firstname = '$delValue'";
$queryLastName = "DELETE FROM quizattempts WHERE lastname = '$delValue'";
$queryStudentId = "DELETE FROM quizattempts WHERE studentid = '$delValue'";
//select which value to search for
if ($_POST["deleteby"] = "attemptid") {
// pass query to database
$result = mysqli_query($conn, $queryAttemptId);
} // end delete attemptid
else if ($_POST["deleteby"] = "firstname") {
// pass query to database
$result = mysqli_query($conn, $queryFirstName);
} // end delete firstname
else if ($_POST["deleteby"] = "lastname") {
// pass query to database
$result = mysqli_query($conn, $queryLastName);
} // end delete lastname
else if ($_POST["deleteby"] = "studentid") {
// pass query to database
$result = mysqli_query($conn, $queryStudentId);
} // end delete student id
echo "this is the result $result";
// if query is successful are found
if ($result) {
echo "<p>Delete operation successful</p>";
} // end if result found
else {
// if no record is found in DB
echo "<p>No records found</p>";
} // end if no result found
} //isset($_POST["attemptid"])
} //$conn
?>