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Php 如何添加提交按钮的操作_Php_Database_Form Submit_Submit Button - Fatal编程技术网

Php 如何添加提交按钮的操作

Php 如何添加提交按钮的操作,php,database,form-submit,submit-button,Php,Database,Form Submit,Submit Button,请回答我的业余问题。我编写了以下代码,用于从输入获取文本并将外接程序添加到数据库: <html xmlns="http://www.w3.org/1999/xhtml" lang="fa"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>My Title</title>

请回答我的业余问题。我编写了以下代码,用于从输入获取文本并将外接程序添加到数据库:

    <html xmlns="http://www.w3.org/1999/xhtml" lang="fa">
    <head>
        <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
        <title>My Title</title>
    </head>
    <body>
        <form action=""  method="POST">
            <label for="name" id="app_name_label"  >name</label>
            <input type="text" name="name" id="name_textfield">

            <input type="submit" name"Submit" value="send" >
        </form>
<?php

if(isset($_POST['Submit']))
{
        include_once("config.php");
    $con = mysql_connect($db_host,$db_user,$db_pass)
    or die(mysql_error());

    $selected=mysql_select_db($db_name, $con) 
    or die(mysql_error());

    if($selected){
    $name = $_REQUEST['name'];

    $ins = "INSERT INTO infos (app_name )  VALUES ('$name')";
    $saved=mysql_query($ins );
    if($saved)
    {
        echo "Saved!!";
    }
    else
    {
        echo "Don't Saved!!";
    }

    }
    mysql_close($con);
}


?>

    </body>
</html>

我的头衔
名称

如果其他一切正常,那么我担心您的代码中有一个简单的输入错误,这会导致您出现问题

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

应该是

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

您在姓名后的代码中错过了等号。:)


另外,当表单方法为POST时,不应使用
$\u REQUEST['name']
,而应使用
$\u POST['name']
。它更安全。其次,尝试使用数据库扩展,而不是使用mysql。从PHP5.5开始,mysql扩展就被弃用。

请参考,最好在正文上方使用PHP标记,并在会话中显示错误,另外,为什么在执行提交后只包含一次(“config.PHP”),而不是在加载实际文件时?这是一个脚本,它与您试图实现的功能相同,但处理效果要好得多

<?php
include_once("config.php");

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

     $err = array();

     $name = $_POST['name'];

     if(!$name){
          $err[] = 'All fields required';
     } 

if(!count($err)){

    $name = mysql_real_escape_string($name); //Make sure you escape unwanted chars

    $row = mysql_fetch_assoc(mysql_query("INSERT INTO infos (app_name )  VALUES ('$name')"));

    if(mysql_affected_rows() == 1)  
    {
        $_SESSION['msg']['succ'] = 'Sent!';
        header("Location: index.php");
    } else{
        $_SESSION['msg']['err'] = 'Noooooo!';
    }   
}

?>

<html xmlns="http://www.w3.org/1999/xhtml" lang="fa">
    <head>
        <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
        <title>My Title</title>
    </head>
    <body>
        <form action=""  method="POST">
            <?php   
            // Lets output message
    if($_SESSION['msg']['err'])
    {
        echo $_SESSION['msg']['err'];
        unset($_SESSION['msg']['err']);
    }
    if($_SESSION['msg']['succ'])
    {
        echo $_SESSION['msg']['succ'];
        unset($_SESSION['msg']['succ']);
    }
        ?>  

            <label id="app_name_label">Name</label>
            <input type="text" name="name" id="name_textfield">

            <input type="submit" name="Submit" value="send">
        </form>
    </body>
</html>

我的头衔
名称

代码应采用两种不同的脚本,首先浏览器请求并显示html部分。然后调用表单
action
属性中记录的脚本,并将表单内容发送到那里。该文件(一个php文件)处理表单数据。代码不必位于两个单独的文件中。问题是相同的:(@Ahmad yeah,有时会发生这种愚蠢的问题:)