PHP cms在文本区域提交不工作_Php_Content Management System_Textarea_Submit

PHP cms在文本区域提交不工作

php content-management-system

PHP cms在文本区域提交不工作,php,content-management-system,textarea,submit,Php,Content Management System,Textarea,Submit,我的php简单cms有问题。我可以在文本区域中获取网站内容，编辑后我想将新编辑的内容发送到数据库，但这不起作用。这是我的表格和功能 editContent.php <html> <head> <title> Basic CMS - Admin Area</title> <h1>Admin Area Edit Content</h1> </head> <bo

我的php简单cms有问题。我可以在文本区域中获取网站内容，编辑后我想将新编辑的内容发送到数据库，但这不起作用。这是我的表格和功能

editContent.php

    <html>
    <head>
    <title> Basic CMS - Admin Area</title>
    <h1>Admin Area Edit Content</h1>
    </head>
    <body>
    <?php
    include ('includes/functions.php');

    $cont = getContent();

    session_start();
    if(isset($_SESSION['user'])) {
    ?>
    <span>Logged In! Welcome <?php echo $_SESSION['user']; ?></span>
    <a href="logout.php">Logout</a>
    <a href="editContent.php">Wijzig content</a>
    <a href="index.php">Admin Home</a>

    <form action="doEditcontent.php" method="post">
    <textarea name="contentarea"><?php echo $cont['content'];?></textarea><br>
    Submit : <input type="submit" value="submit" />
    </form>

<?php
} else {
    header("Location: login.php");
}
?>
</body>
</html>

<?php
include('includes/functions.php');

if(isset($_POST['submit'])) {
    if(isset($_POST['contentarea'])){       
        editContent($_POST['contentarea']);
        header("Location: ../index.php?page=2");   
    } else 
        echo "Please enter some content!";
} else {
    header("Location: ../index.php?page=1");
}
?>


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提交：

functions.php

<?php
include('includes/connect.php');

function getContent(){
    $query = mysql_query("SELECT content FROM taalcontent WHERE taalid = 1 AND contentid = 1") or die (mysql_error());
    return mysql_fetch_assoc($query);
    echo $query;
}

function editContent($pContent) {
    if(isset($pContent)){
        $query = "UPDATE taalcontent SET content content = '$pContent'  WHERE contendid = 1 AND taalid = 1";
    } else {
        echo "fout";
    }
    mysql_query($query);
}

尝试更改
editContent($_GET['content']);

到
实际上，您要求它将内容设置为？get
参数中的任何内容（我看不到其他地方引用了该参数，因此我猜它没有填充任何内容）
：）调用函数editContent（$pContent）时传递的参数错误。此外，您应该在引用函数之前定义它。
试着这样做：
    <form action="doEditcontent.php" method="post">
    <textarea name="contentarea"><?php echo $cont['content'];?></textarea><br>
    Submit : <input type="submit" value="submit" />
    </form>

    <?php
    include('includes/functions.php');


    function editContent($pContent) {
if(isset($pContent)){
    $query = "UPDATE taalcontent SET content content = '$pContent'  WHERE contendid = 1 AND taalid = 1";
} else {
    echo "fout";
}
mysql_query($query);
    }


    if(isset($_POST['submit'])) {
if(isset($_POST['contentarea'])){       
    editContent($_POST['contentarea']);
    header("Location: ../index.php?page=2");   
} else 
    echo "Please enter some content!";
    } else {
header("Location: ../index.php?page=1");
    }
    ?>




提交：
这些只是我从不同文件中得到的代码片段。我在页面顶部包含了它们。我在调用函数editContent之前定义了该函数，并且按照Dom的建议，将$_POST['contentarea']而不是$_GET['content']）。我现在发布了我的所有文件，也许你现在更清楚了
    <form action="doEditcontent.php" method="post">
    <textarea name="contentarea"><?php echo $cont['content'];?></textarea><br>
    Submit : <input type="submit" value="submit" />
    </form>

    <?php
    include('includes/functions.php');


    function editContent($pContent) {
if(isset($pContent)){
    $query = "UPDATE taalcontent SET content content = '$pContent'  WHERE contendid = 1 AND taalid = 1";
} else {
    echo "fout";
}
mysql_query($query);
    }


    if(isset($_POST['submit'])) {
if(isset($_POST['contentarea'])){       
    editContent($_POST['contentarea']);
    header("Location: ../index.php?page=2");   
} else 
    echo "Please enter some content!";
    } else {
header("Location: ../index.php?page=1");
    }
    ?>