[英]Using POST to insert data into a database
我需要编写代码以将联系人添加到数据库。 当我运行addcontact.html表单时,会将新行添加到数据库中,但是唯一包含任何数据的字段是自动编号字段。 其余为空白。 我知道我的代码中肯定有一个错误,但是我似乎找不到它在哪里。
这是我的insert.php
<?php
$con=mysqli_connect("localhost","root","","mycontacts");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$name = mysqli_real_escape_string($con, $_POST['name']);
$address = mysqli_real_escape_string($con, $_POST['address']);
$city = mysqli_real_escape_string($con, $_POST['city']);
$state = mysqli_real_escape_string($con, $_POST['state']);
$zip = mysqli_real_escape_string($con, $_POST['zip']);
$phone = mysqli_real_escape_string($con, $_POST['phone'])
$sql="INSERT INTO Contacts (contac_id, contact_name, contact_address, contact_city,
contact_state, contact_zip_code, contact_phones, contact_website)
VALUES ('', '$name', '$address', '$city', '$state', '$zip',
'$phone', '')";
if (!mysqli_query($con,$sql)) {
die('Error: ' . mysqli_error($con));
}
header('Location:mainmenu.php');
session_start();
$_SESSION['loggedIn'] = true;
mysqli_close($con);
?>
这是我在addcontact.html上的html表单的代码:
<form action="insert.php" method='post'>
<table width="250">
<tr>
<p>
<td align="right" valign="middle">
Name:
</td>
<td align="right" valign="middle">
<input type="text" class= "textbox" name="name" />
</td>
</p>
</tr>
<tr>
<p>
<td align="right" valign="middle">
Address:
</td>
<td align="right" valign="middle">
<input type="text" class= "textbox" name="address" />
</td>
</p>
</tr>
<tr>
<p>
<td align="right" valign="middle">
City:
</td>
<td align="right" valign="middle">
<input type="text" class= "textbox" name="city" />
</td>
</p>
</tr>
<tr>
<p>
<td align="right" valign="middle">
State:
</td>
<td align="right" valign="middle">
<input type="text" class= "textbox" name="state" />
</td>
</p>
<tr>
<p>
<td align="right" valign="middle">
Zip:
</td>
<td align="right" valign="middle">
<input type="text" class= "textbox" name="zip" maxlength='10'/>
</td>
</p>
</tr>
<tr>
<p>
<td align="right" valign="middle">
Phone
</td>
<td align="right" valign="middle">
<input type="text" class= "textbox" name="phone" />
</td>
</p>
</tr>
<tr>
<td colspan="2">
<center>
<br>
<input type='submit' class= "submitbutton" value="Submit"/>
我确实回显了我的$ sql,这是我得到的错误:
( ! ) Notice: Undefined index: name in D:\wamp\www\it665\insert.php on line 7
( ! ) Notice: Undefined index: address in D:\wamp\www\it665\insert.php on line 8
( ! ) Notice: Undefined index: city in D:\wamp\www\it665\insert.php on line 9
( ! ) Notice: Undefined index: state in D:\wamp\www\it665\insert.php on line 10
( ! ) Notice: Undefined index: zip in D:\wamp\www\it665\insert.php on line 11
( ! ) Notice: Undefined index: phone in D:\wamp\www\it665\insert.php on line 12
INSERT INTO Contacts (contac_id, contact_name, contact_address, contact_city,
contact_state, contact_zip_code, contact_phones, contact_website) VALUES ('', '', '',
'', '', '', '', '')
因此,然后我取出了echo $ sql,并确保删除了标头代码,这样它就不会转发给我,并且错误仍然存在。 所以我至少要到某个地方。
Undefined index: address in D:\\wamp\\www\\it665\\insert.php on line 8
$_POST['address']
表示您没有$_POST['address']
。 您没有发送您认为自己的POST数据。
在$_POST
数组上执行var_dump
,然后查看其中的内容。
您可以尝试此操作(如果contac_id是自动中断的)
$sql="INSERT INTO Contacts (contact_name, contact_address, contact_city,
contact_state, contact_zip_code, contact_phones, contact_website)
VALUES ('$name', '$address', '$city', '$state', '$zip',
'$phone', '')";
代替
$sql="INSERT INTO Contacts (contac_id, contact_name, contact_address, contact_city,
contact_state, contact_zip_code, contact_phones, contact_website)
VALUES ('', '$name', '$address', '$city', '$state', '$zip',
'$phone', '')";
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