Php 找不到我的Url地址或我的编程有问题
我已经创建了一个数据库(在phpmyadmin上)和一个使用php和html的表单,但我似乎找不到我的数据库URL地址,或者我的表单有一些非常错误的地方。。。它说的是Error-1-Bridge响应Error,请检查API文档或这个ajax响应。这是什么意思? 代码如下:Php 找不到我的Url地址或我的编程有问题,php,html,Php,Html,我已经创建了一个数据库(在phpmyadmin上)和一个使用php和html的表单,但我似乎找不到我的数据库URL地址,或者我的表单有一些非常错误的地方。。。它说的是Error-1-Bridge响应Error,请检查API文档或这个ajax响应。这是什么意思? 代码如下: <html> <head> <title>Database</title> </head> <body> &
<html>
<head>
<title>Database</title>
</head>
<body>
<?php
if(isset($_POST['update'])) {
$dbhost = localhost;'localhost:id674442_wommath';
$dbuser = 'root';
$dbpass = 'passroot';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn ) {
die('Could not connect: ' . mysql_error());
}
$fir_name = $_POST['fir_name'];
$sur_name = $_POST['sur_name'];
$li_points = $_POST['li_points'];
$xp_points = $_POST['xp_points'];
$sql = "UPDATE First_name ". "SET Sur_name = $sur_name ".
"WHERE fir_name= $fur_name" ;
mysql_select_db('test_db');
$retval = mysql_query( $sql, $conn );
if(! $retval ) {
die('Could not update data: ' . mysql_error());
}
echo "Updated data successfully\n";
mysql_close($conn);
}else {
?>
<form method = "post" action = "<?php $_PHP_SELF ?>">
<table width = "400" border =" 0" cellspacing = "1"
cellpadding = "2">
<tr>
<td width = "100">First_Name</td>
<td><input name = "fir_name" type = "text"
id = "emp_id"></td>
</tr>
<tr>
<td width = "100">Surname</td>
<td><input name = "sur_name" type = "text"
id = "emp_id"></td>
</tr>
<tr>
<td width = "100">life_points</td>
<td><input name = "li_points" type = "text"
id = "emp_id"></td>
</tr>
<tr>
<td width = "100">xp_points</td>
<td><input name = "xp_points" type = "text"
id = "emp_salary"></td>
</tr>
<tr>
<td width = "100"> </td>
<td> </td>
</tr>
<tr>
<td width = "100"> </td>
<td>
<input name = "update" type = "submit"
id = "update" value = "Update">
</td>
</tr>
</table>
</form>
<?php
}
?>
</body>
</html>
数据库
检查这条线
$dbhost = localhost;'localhost:id674442_wommath';
为什么有分号
将$dbhost
设置为“localhost”
,它就会工作。问题在于$dbhost
参数
您必须将其设置为正确的值
如果你能分享更多的细节,帮助你会很容易
如果您正在使用localhost
use,则必须编写如下内容:
$dbhost='localhost'
或$dbhost='127.0.0.1'代码>
并确保您的Apache服务器正在使用端口80,如果端口80不同,则必须在localhost或ip之后写入端口,如下所示:
$dbhost='localhost:port number'
或$dbhost='127.0.0.1:端口号'代码>我已经更新了代码,代码中有很多错误。另外,如果您使用的是php7,那么一定要使用mysqli,而不仅仅是mysql
<html>
<head>
<title>Database</title>
</head>
<body>
<?php
if(isset($_POST['update'])) {
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = 'passroot';
$conn = mysqli_connect($dbhost, $dbuser, $dbpass);
if(! $conn ) {
die('Could not connect: ' . mysqli_error());
}
$fir_name = $_POST['fir_name'];
$sur_name = $_POST['sur_name'];
$li_points = $_POST['li_points'];
$xp_points = $_POST['xp_points'];
$sql = "UPDATE First_name SET Sur_name = '".$sur_name."' WHERE fir_name= '".$fir_name."'";
mysqli_select_db('test_db');
$retval = mysqli_query( $sql, $conn );
if(! $retval ) {
die('Could not update data: ' . mysqli_error());
}
echo "Updated data successfully\n";
mysqli_close($conn);
}
else {
?>
<form method = "post" action = "<?php $_PHP_SELF ?>">
<table width = "400" border =" 0" cellspacing = "1"
cellpadding = "2">
<tr>
<td width = "100">First_Name</td>
<td><input name = "fir_name" type = "text"
id = "emp_id"></td>
</tr>
<tr>
<td width = "100">Surname</td>
<td><input name = "sur_name" type = "text"
id = "emp_id"></td>
</tr>
<tr>
<td width = "100">life_points</td>
<td><input name = "li_points" type = "text"
id = "emp_id"></td>
</tr>
<tr>
<td width = "100">xp_points</td>
<td><input name = "xp_points" type = "text"
id = "emp_salary"></td>
</tr>
<tr>
<td width = "100"> </td>
<td> </td>
</tr>
<tr>
<td width = "100"> </td>
<td>
<input name = "update" type = "submit"
id = "update" value = "Update">
</td>
</tr>
</table>
</form>
<?php
}
?>
</body>
</html>
数据库
对不起,到底是什么问题?我的问题是。。。我找不到我的URL地址,或者我的html或php错误。所以,当我填写表格并按update时,我会说找不到Url地址。但我确信我的URL地址是正确的。我已将代码设置为$dbhost=“localhost”:“localhost:id674442_wommath”;但它仍然不起作用…它仍然不起作用。。。当我按update时,它仍显示此信息。找不到此服务器上找不到请求的URL/<。但我确信我的URL地址是正确的。在html格式中,更新操作,你希望我把它放在哪里?因为它没有改变什么-(您的问题已解决..接受答案并结束此问题