我试图在将注册表单的所有内容放入数据库之前测试对 MySQL 数据库的写入，但这是错误的。有人可以检查一下我的 PHP 代码，看看我哪里出错了。<html> <body> <form action="testingPHP.php" method="post"> Name: <input type="text" name="Name"><br> Age: <input type="text" name="Age"><br> Gender: <select name="Gender"> <option value=" "> </option> <option value="Boy">Boy</option> <option value="Girl">Girl</option> </select><br> <input type ="submit"> </form> </body></html> PHP：$Name = $_.POST['Name'];$Age = $_.POST['Age'];$Gender = $_.POST['Gender'];$servername = "localhost";$username = "testing";$password = "testing1";$db = "tesingdb";$conn new mysql($sername, $username, $password, $db);$sql = "insert into testingtable (name,age,gender)values('$Name','$Age','$Gender')"$conn->close();
1 回答
素胚勾勒不出你
TA贡献1827条经验 获得超9个赞
以下是变化。
已删除
testingPHP.php自<form action="testingPHP.php" ...>$_.POST到$_POST$conn = new mysqli(....)代替$conn new mysql(....)添加
$conn->query($sql);拼写错误
$servername
其余代码
<form action="" method="post">
Name: <input type="text" name="Name"><br>
Age: <input type="text" name="Age"><br>
Gender: <select name="Gender">
<option value=" "> </option>
<option value="Boy">Boy</option>
<option value="Girl">Girl</option>
</select><br>
<input type ="submit">
</form>
<?php
$Name = $_POST['Name'];
$Age = $_POST['Age'];
$Gender = $_POST['Gender'];
$servername = "localhost";
$username = "testing";
$password = "testing1";
$db = "tesingdb";
$conn = new mysqli($servername, $username, $password, $db);
$sql = "insert into testingtable (name,age,gender) values('$Name','$Age','$Gender')";
$conn->query($sql);
$conn->close();
?>
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