我正在尝试从这样的网址获取ID :( 37是ID)
http://localhost/myQSP/public/assign-qsps.php?employeeName=37
并将其与复选框选择一起插入数据库。
数据库
id | employees_fk | qsps_fk | checked
1 37 101 1
2 37 102 1
qsps_fk是一个关系表。使用此流程代码
<?php
if (isset($_POST['submit'])) {
$assign_qsps_id = $_POST["hidden-qsp-checkBoxes"];
$hidden_qsp = $_POST["department-qsp-checkBoxes"];
if(isset($_POST['hidden-qsp-checkBoxes'])) {
foreach ($assign_qsps_id as $qsp_id){
$visible_assign_qsp = in_array($qsp_id, $hidden_qsp) ? 1 : 0;
$query = "INSERT INTO junction_employees_qsps (qsps_fk, checked) VALUES ($qsp_id, $visible_assign_qsp);
$result = mysqli_query($db_connection, $query);
}
}
}
?>
它会创建一个像这样的表
id | employees_fk | qsps_fk | checked
1 0 101 1
2 0 102 1
我的表单是
<form action="assign-qsps-process.php" method="post">
<input type="submit" name="submit" value="ASSIGN QSP'S" />
<ul>
<?php $employee_qsps = find_departments_for_assign_qsps($current_employee['id']);
while($qsp = mysqli_fetch_assoc($employee_qsps)) { ?>
<li>
<input type='checkbox' name='hidden-qsp-checkBoxes[]' value='<?php htmlentities($qsp["qsp_id"]); ?>' style='display:none' checked='checked' />";
<input type='checkbox' name='department-qsp-checkBoxes[]' value='<?php htmlentities($qsp["qsp_id"]); ?>' />
<?php echo htmlentities($qsp['qsp_name']); ?>
</li>
}
</ul>
</form>
那么如何从带有复选框选择的URL中获取ID以及使employees_fk成为关系表?
我已经尝试了这个并且它不起作用。
<?php
if (isset($_POST['submit'])) {
$id_employee = $current_employee["id"];
$assign_qsps_id = $_POST["hidden-qsp-checkBoxes"];
$hidden_qsp = $_POST["department-qsp-checkBoxes"];
if(isset($_POST['hidden-qsp-checkBoxes'])) {
foreach ($assign_qsps_id as $qsp_id){
$visible_assign_qsp = in_array($qsp_id, $hidden_qsp) ? 1 : 0;
$query = "INSERT INTO junction_employees_qsps (employees_fk, qsps_fk, checked) VALUES ($id_employee, $qsp_id, $visible_assign_qsp);
$result = mysqli_query($db_connection, $query);
}
}
}
?>
谢谢!我非常感谢帮助!