下面的代码不起作用

Question

这是我的表单代码，我希望用户输入的数据应该保存在 mysql...使用 wamp..任何解决方案 plzz..请重写代码并粘贴它。 ..................................................... ...................................

</head>
<body>

<form action="form.php" method="post">

    Project_Manager:<input type="text" name="ProjectManager">
<br />

    Email_Address:<input type="text" name="EmailAddress">
<br />

    Company_Name:<input type="text" name="CompanyName">
<br />

    Company_Street_Address:<input type="text" name="CompanyStreetAddress">
<br />

    Company_Suit:<input type="text" name="CompanySuit">
<br />

    City:<input type="text" name="City">
<br />

    State:<input type="text" name="State">
<br />

    Zip_code:<input type="text" name="Zipcode">
<br />

    Company_Phone:<input type="text" name="CompanyPhone">
<br />

    Billing_Name:<input type="text" name="BillingName">
<br />

    Billing_Company_Street_Address:<input type="text" name="BillingCompanyStreetAddress">
<br />

    Billing_Company_Suite:<input type="text" name="BillingCompanySuite">
<br />

    Billing_City:<input type="text" name="BillingCity">
<br />

    Billing_State:<input type="text" name="BillingState">
<br />

    Billing_Zip_Code:<input type="text" name="BillingZipCode">
<br />

    Billing_Phone:<input type="text" name="BillingPhone">
<br />


    <input type="submit" value="submit">

    </form>



<?php

    if(isset($_POST['submit'])){

    $con = mysql_connect("localhost","root","");
    if(!$con){
         die('Could not connect: ' . mysql_error());
    }

mysql_select_db("ecc1",$con);   


$sql = "INSERT INTO client(
    Project_Manager,
    Email_Address, 
    Company_Name, 
    Company_Street_Address, 
    Company_Suit, 
    City, 
    State, 
    Zip_code, 
    Company_Phone, 
    Billing_Name, 
    Billing_Company_Street_Address, 
    Billing_Company_Suite, 
    Billing_City, 
    Billing_State, 
    Billing_Zip_Code, 
    Billing_Phone
)
    VALUES( '$_POST[ProjectManager]',
            '$_POST[EmailAddress]',
            '$_POST[CompanyName]',
            '$_POST[CompanyStreetAddress]',
            '$_POST[CompanySuit]',
            '$_POST[City]',
            '$_POST[State]',
            '$_POST[Zipcode]',
            '$_POST[CompanyPhone]',
            '$_POST[BillingName]',
            '$_POST[BillingCompanyStreetAddress]',
            '$_POST[BillingCompanySuite]',
            '$_POST[BillingCity ]',
            '$_POST[BillingState]',
            '$_POST[BillingZipCode]',
            '$_POST[BillingPhone]',
            )";

mysql_query($sql,$con);
mysql_close($con);
    }

?>

</body>

</html>

Answer 1

很可能您对表单是否已提交的测试失败了：

if(isset($_POST['submit'])){
    // this is probably false
}

如果您将 name="submit" 添加到您的提交按钮，您可能会得到您期望的结果。

<input type="submit" name="submit" value="submit">

Answer 2

好吧，对于初学者来说，您使用的是已弃用的 mysql_* 扩展，它将不再适用于 PHP 的新版本，其次，您在查询数据库时使用 MySQLi 标准。

您的整个脚本应该如下所示：

HTML:

<input type="submit" value="true" name="submit"> <!-- set the name of submit button to "submit" so your if statement works !-->

PHP:

$conn = new mysqli('localhost', 'root', '', 'ecc1'); // use mysqli instead of mysql. mysql is deprecated

if(isset($_POST['submit'])){
    $sql = "INSERT INTO client(
        Project_Manager,
        Email_Address, 
        Company_Name, 
        Company_Street_Address, 
        Company_Suit, 
        City, 
        State, 
        Zip_code, 
        Company_Phone, 
        Billing_Name, 
        Billing_Company_Street_Address, 
        Billing_Company_Suite, 
        Billing_City, 
        Billing_State, 
        Billing_Zip_Code, 
        Billing_Phone) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)"; // your sql
    $stmt = $conn->prepare($sql); // to protect against sql injection
    $stmt->bind_param("ssssssiisssssii", $_POST['ProjectManager'], $_POST'[EmailAddress'], $_POST['CompanyName'], $_POST['CompanyStreetAddress'], $_POST['CompanySuit'], $_POST['City'], $_POST['State'], $_POST['Zipcode'], $_POST['CompanyPhone'],  $_POST['BillingName'], $_POST['BillingCompanyStreetAddress'], $_POST['BillingCompanySuite'], $_POST['BillingCity'], $_POST['BillingState'], $_POST['BillingZipCode'], $_POST['BillingPhone']); // set all the values of the row columns

    if($stmt->execute()){ // if the query is successful
        die("Query Successful"); // output success message
    }
}

现在，假设我已正确完成所有操作（包括计算出您拥有的大量 post 数据），查询应该执行，您应该得到留言：

Query Successful

请注意：对于"ssssssiisssssii"，s表示字符串，i表示整数。

如果出现任何问题，或者您遇到任何错误，请在评论中留下。

要了解更多关于预处理语句和 MySQLi 本身的信息，请查看 PHP 手册：http://php.net/manual/en/mysqli.prepare.php

Answer 3

<?php

$sql = "INSERT INTO client(
    Project_Manager,
    Email_Address, 
    Company_Name, 
    Company_Street_Address, 
    Company_Suit, 
    City, 
    State, 
    Zip_code, 
    Company_Phone, 
    Billing_Name, 
    Billing_Company_Street_Address, 
    Billing_Company_Suite, 
    Billing_City, 
    Billing_State, 
    Billing_Zip_Code, 
    Billing_Phone
)
    VALUES( '{$_POST['ProjectManager']}',
            '{$_POST['EmailAddress']}',
            '{$_POST['CompanyName']}',
            '{$_POST['CompanyStreetAddress']}',
            '{$_POST['CompanySuit']}',
            '{$_POST['City']}',
            '{$_POST['State']}',
            '{$_POST['Zipcode']}',
            '{$_POST['CompanyPhone']}',
            '{$_POST['BillingName']}',
            '{$_POST['BillingCompanyStreetAddress']}',
            '{$_POST['BillingCompanySuite']}',
            '{$_POST['BillingCity']}',
            '{$_POST['BillingState']}',
            '{$_POST['BillingZipCode']}',
            '{$_POST['BillingPhone']}',
            )";

mysql_query($sql,$con);
mysql_close($con);
    }

?>

下面的代码不起作用

the below code is not working

php

html

wampserver