我有一个PHP页面,它填充了我数据库的2个不同的表。
添加新列然后创建外键后,无法再在该表中插入数据。如果我删除了外键,那么它再次起作用......之前有人经历过这样的事情吗?
这是我从同一页面填充两个表格的方式:
include("../includes/connection.php");
$name = mysqli_real_escape_string($link, $_POST['name']);
$email = mysqli_real_escape_string($link, $_POST['email']);
$number = mysqli_real_escape_string($link, $_POST['number']);
$device = mysqli_real_escape_string($link, $_POST['device']);
$price = mysqli_real_escape_string($link, $_POST['price']);
$payment = mysqli_real_escape_string($link, $_POST['payment']);
$status = mysqli_real_escape_string($link, $_POST['status']);
$model = mysqli_real_escape_string($link, $_POST['model']);
$problem = mysqli_real_escape_string($link, $_POST['problem']);
// attempt insert query execution
$sql = "INSERT INTO customer (name, mail, number, device, price, paymenttype,status,date) VALUES ('$name', '$email', '$number', '$device', '$price', '$payment','$status',NOW())";
if(mysqli_query($link, $sql)){
// echo "Records added successfully.";
header("location:add-customer.php?message=The customer has been added to the database1");
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
$sql = "INSERT INTO job (device, model, problem, status) VALUES ('$device', '$model', '$problem', '$status')";
if(mysqli_query($link, $sql)){
// echo "Records added successfully.";
header("location:add-customer.php?message=The customer has been added to the database2");
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
} mysqli_close($link);?>
而且,这张照片可能更好地解释了我想要实现的目标:
例:
按钮"信息"在第一行,想要得到客户表的第一行..
按钮"信息"在第二行,想要获得第二行客户表...
---------模式弹出代码,点击-----
后收集客户表<div class="modal fade" id="myModal" role="dialog">
<div class="modal-dialog modal-lg">
<!-- Modal content-->
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal">×</button>
<h4 class="modal-title">Customer Information</h4>
</div>
<div class="modal-body">
<?php
include("../includes/connection.php");
if ($link->connect_errno > 0) {
die('Unable to connect to database [' . $link->connect_error . ']');
}
$sql = "SELECT id,name,mail,number,price,paymenttype,faktura,date from customer WHERE id = '[job_id]' ";
if (!$result = $link->query($sql)) {
die('There was an error running the query [' . $link->error . ']');
}
echo "
<table class='table'>
<thead>
<tr>";
/* Get field information for all columns */
while ($finfo = $result->fetch_field()) {
echo "
<th>" . $finfo->name . "</th>";
}
echo "
</tr>
</thead>
<tbody>";
while ($row = $result->fetch_assoc()) {
echo "<tr class='info'>
<td>" . $row['id'] . "</td>
<td>" . $row['name'] . "</td>
<td>" . $row['mail'] . "</td>
<td>" . $row['number'] . "</td>
<td>" . $row['price'] . "</td>
<td>" . $row['paymenttype'] . "</td>
<td>" . $row['faktura'] . "</td>
<td>" . $row['date'] . "</td>
</tr>";
}
echo "
</tbody>
</table>";
?>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
</div>
</div>
</div>
</div>
---------并使用此代码收集工作表----
<?php
include("../includes/connection.php");
if ($link->connect_errno > 0) {
die('Unable to connect to database [' . $link->connect_error . ']');
}
if (isset($_POST['update'])) {
$results = $link->query("UPDATE job SET status='$_POST[status]', priority='$_POST[priority]' WHERE id='$_POST[hidden]'");
$results = $link->query("UPDATE customer SET status='$_POST[status]' WHERE id='$_POST[hidden]'");
}
$sql = "SELECT * from job";
if (!$result = $link->query($sql)) {
die('There was an error running the query [' . $link->error . ']');
}
echo "
<table class='table'>
<thead>
<tr>";
/* Get field information for all columns */
while ($finfo = $result->fetch_field()) {
echo "
<th>" . $finfo->name . "</th>";
}
echo "
</tr>
</thead>
<tbody>";
while ($row = $result->fetch_assoc()) {
$job_id = $row['id'];
echo "<form action='' method=post>";
echo "<tr class='info'>
<input type=hidden name=hidden value=" . $row['id'] . ">
<td>" . $row['id'] . "</td>
<td>" . $row['device'] . "</td>
<td>" . $row['model'] . "</td>
<td>" . $row['problem'] . "</td>
<td>
<select class='form-control col-sm-10' id='status' name='status'>
<option value='new' ". ($row['status'] == 'new'? 'selected ': '') .">New</option>
<option value='progress' ". ($row['status'] == 'progress'? 'selected ': '') .">Progress</option>
<option value='wait' ". ($row['status'] == 'wait'? 'selected ': '') .">Wait</option>
<option value='done' ". ($row['status'] == 'done'? 'selected ': '') .">Done</option>
<option value='close' ". ($row['status'] == 'close'? 'selected ': '') .">Close</option>
</select>
</td>
<td><select class='form-control col-sm-10' id='priority' name='priority'>
<option style='background-color:green;'value='low' ". ($row['priority'] == 'Low'? 'selected ': '') .">Low</option>
<option style='background-color:yellow; value='Medium' ". ($row['priority'] == 'Medium'? 'selected ': '') .">Medium</option>
<option style='background-color:red; value='High' ". ($row['priority'] == 'High'? 'selected ': '') .">High</option>
</select></td>
<td> <button type='submit' class='btn btn-primary btn-sm' name='update'>Update</button></td>
<td> <a class='btn btn-primary btn-sm' data-toggle='modal' data-target='#myModal' name='job_id' value='[$job_id]' > Info</a></td>
</tr>";
echo "</form>";
}
echo "
</tbody>
</table>";
?>
答案 0 :(得分:1)
这个脚本似乎有很多错误。
由于首先插入了echo "<table class=box border=1px>
<div id=<wrapper>
<tr>
<th>Zone</th>
<th>Rownumber</th>
</tr>";
echo "<form action=selectsits.php enctype=multipart/form-data method=post>";
while($record=mysql_fetch_array($availablesitsdata) ) {
echo '<tr>';
echo '<select name=cat_id>
<option value="a">aa</option>
<option value="b">bb</option>
</select>
</td>';
echo "<td>" . $record['Zone'] . "</td>";
echo"</tr>";
}
echo "</form>";
表行,并且未指定Customer
,我认为它正在尝试使用0,这会使外键约束失败。
我无法从您的表中了解哪个实体需要先添加,或者哪个实体依赖于另一个实体。所以我会给你一个适合两者的解决方案。我会说,虽然这个解决方案有效,它允许你忽略FK约束,如果你打算忽略它,那么你最好不要首先设置它。
首先更改job_id
表格上的job_id
字段以允许Customer
值。
第二次更改此行
NULL
到这一行
$sql = "INSERT INTO customer (name, mail, number, device, price, paymenttype,status,date) VALUES ('$name', '$email', '$number', '$device', '$price', '$payment','$status',NOW())";
似乎在执行期间可以访问两个标头重定向。
您应始终在标题重定向后调用$sql = "INSERT INTO customer (name, mail, number, device, price, paymenttype,status,date,job_id) VALUES ('$name', '$email', '$number', '$device', '$price', '$payment','$status',NOW(),NULL)";
,否则脚本将继续处理。 Why I have to call 'exit' after redirection through header('Location..') in PHP?
如果您回显错误消息并且您没有缓冲输出,那么您的重定向将失败,因为您已经发送了输出,它会发送标头。 How to fix "Headers already sent" error in PHP
exit()
Edit after problem clarification