无法在数据库中输入数据
无法在数据库中输入数据
提问于 2016-10-31 13:06:32
<html>
<head>
<title>Add New Record in MySQL Database</title>
</head>
<body>
<?php
if(isset($_POST['add'])) {
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = 'password';
$conn = mysqli_connect($dbhost, $dbuser, $dbpass);
if(! $conn ) {
die('Could not connect: ' . mysql_error());
}
if(! get_magic_quotes_gpc() ) {
$emp_name = addslashes ($_POST['emp_name']);
$emp_address = addslashes ($_POST['emp_address']);
}else {
$emp_name = $_POST['emp_name'];
$emp_address = $_POST['emp_address'];
}
$emp_salary = $_POST['emp_salary'];
$sql = "insert into employee(emp_name,emp_address, emp_salary)values('$emp_name','$emp_address','$emp_salary')";
mysqli_select_db($conn,"test_db");
$retval = mysqli_query($conn,$sql);
if(!$retval) {
die('Could not enter data: ' . mysql_error());
}
echo "Entered data successfully\n";
mysql_close($conn);
}else {
?>
<form method = "post" action = "<?php $_PHP_SELF ?>">
<table width = "400" border = "0" cellspacing = "1"
cellpadding = "2">
<tr>
<td width = "100">Employee Name</td>
<td><input name = "emp_name" type = "text"
id = "emp_name"></td>
</tr>
<tr>
<td width = "100">Employee Address</td>
<td><input name = "emp_address" type = "text"
id = "emp_address"></td>
</tr>
<tr>
<td width = "100">Employee Salary</td>
<td><input name = "emp_salary" type = "text"
id = "emp_salary"></td>
</tr>
<tr>
<td width = "100"> </td>
<td> </td>
</tr>
<tr>
<td width = "100"> </td>
<td>
<input name = "add" type = "submit" id = "add"
value = "Add Employee">
</td>
</tr>
</table>
</form>
<?php
}
?>当我试图输入值并在此时按submit按钮时，我不会收到任何错误，但是我无法在数据库中输入值。问题是我收到的短信是“无法输入数据:表‘雇员’是只读的” 我已经在wamp服务器中创建了数据库(test_db)和表(employee )。
回答 2
Stack Overflow用户
回答已采纳
发布于 2016-10-31 13:45:15
你的问题解决了。尽管如此，我仍然强烈建议您使用已准备好的语句，否则您的代码将被打开，用于SQL注入和可能出现的引用问题。
你把mysql和mysqli混在一起。别说了。由于您正在使用mysqli，所以要利用准备好的语句和bind_param，否则您将面临Since和可能的引用问题。- @aynber
改变
- 将
die('Could not connect: ' . mysql_error());更改为die('Could not connect: ' . mysqli_connect_error()); - 将
mysql_close($conn);更改为mysqli_close($conn); - 将
action = "<?php $_PHP_SELF ?>"更改为action = "<?php echo $_SERVER['PHP_SELF']; ?>" - 使用准备好的陈述。
更新代码
<html>
<head>
<title>Add New Record in MySQL Database</title>
</head>
<body>
<?php
if(isset($_POST['add'])) {
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = 'password';
$db = "test_db";
$conn = mysqli_connect($dbhost, $dbuser, $dbpass, $db);
if(! $conn ) {
die('Could not connect: ' . mysqli_connect_error());
}
$stmt = mysqli_prepare($conn, "INSERT INTO employee(emp_name,emp_address, emp_salary) VALUES (?, ?, ?)");
mysqli_stmt_bind_param($stmt, 'sss', $_POST['emp_name'], $_POST['emp_address'], $_POST['emp_salary']);
if(!mysqli_stmt_execute($stmt)) {
die('Could not enter data: ' . mysqli_error($conn));
}
echo "Entered data successfully\n";
mysqli_close($conn);
} else {
?>
<form method = "post" action = "<?php echo $_SERVER['PHP_SELF']; ?>">
<table width = "400" border = "0" cellspacing = "1" cellpadding = "2">
<tr>
<td width = "100">Employee Name</td>
<td><input name = "emp_name" type = "text" id = "emp_name"></td>
</tr>
<tr>
<td width = "100">Employee Address</td>
<td><input name = "emp_address" type = "text" id = "emp_address"></td>
</tr>
<tr>
<td width = "100">Employee Salary</td>
<td><input name = "emp_salary" type = "text" id = "emp_salary"></td>
</tr>
<tr>
<td width = "100"> </td>
<td> </td>
</tr>
<tr>
<td width = "100"> </td>
<td><input name = "add" type = "submit" id = "add" value = "Add Employee"></td>
</tr>
</table>
</form>
<?php
}
?>快速查找
Stack Overflow用户
发布于 2016-10-31 13:26:49
我确信您的用户没有被授予向您的表中输入数据的权限--请编辑schema_name，并对您的DB执行查询：
将表schema_name.employee上的所有内容授予根；
您也可以尝试不使用架构:将表employee上的所有内容都授予root；
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/40342806
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