<form action="" method="post">
<input type="text" id="title" name="title" />
<input type="text" id="link" name="link" />
<input type="submit" value="Add resource" />
<?php
if(isset($_POST['title']) && $_POST['link']) {
$t = $_POST['title'];
$l = $_POST['link'];
$con = mysqli_connect("localhost","root","","rman");
if (mysqli_connect_errno()) {
die("Failed to connect to MySQL:" . mysqli_connect_error());
}
mysqli_query($con, "INSERT INTO tutorials (id, title, link, section) VALUES ('','$t','$l','')");
}
?>
</form>
这怎么可能不工作?我已经切除了所有可能导致这一切的部分。数据库中没有任何内容,也没有任何错误返回。
每个人都想知道:DB名称:rman表名:教程列:id (INT11, Auto increment), title (Text), link (Text), section(INT11)
我是瞎了吗?如果是这样的话,我很抱歉。希望有人能看到我做错了什么,帮助我。
应该可以。
<?php
if(isset($_POST['title']) && $_POST['link']) {
$t = $_POST['title'];
$l = $_POST['link'];
$con = mysqli_connect("localhost","root","","rman");
if (mysqli_connect_errno()) {
die("Failed to connect to MySQL:" . mysqli_connect_error());
}
mysqli_query($con, "INSERT INTO tutorials (id, title, link, section) VALUES ('', '$t', '$l', '1')");
}
?>
<form action="" method="post">
<input type="text" id="title" name="title" />
<input type="text" id="link" name="link" />
<input type="submit" name="submit" value="Add resource" />
</form>
使用mysqli_error()检查错误代码,以及不插入空白id,如果它是AI
:
if (!mysqli_query($con, "INSERT INTO tutorials (id, title, link, section) VALUES ('','$t','$l','')"))
{
echo mysqli_error($con);
}