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PHP:从if语句中检索值,以便在同一页面上的html输入标记中使用_Php_Html_If Statement_Scope - Fatal编程技术网

PHP:从if语句中检索值,以便在同一页面上的html输入标记中使用

PHP:从if语句中检索值,以便在同一页面上的html输入标记中使用,php,html,if-statement,scope,Php,Html,If Statement,Scope,我试图从if语句中检索值,也就是从数据库中获取值,这是我的代码 <!DOCTYPE html> <?PHP include('session.php'); include('recipedbconfig.php'); if(isset($_POST['submit'])) { $postrcpnmesrch = $_POST['rcpnmesrch']; if($_SERVER["REQUEST_METHOD"] =

我试图从if语句中检索值,也就是从数据库中获取值,这是我的代码

<!DOCTYPE html>
<?PHP
    include('session.php');
    include('recipedbconfig.php');


    if(isset($_POST['submit'])) {
        $postrcpnmesrch = $_POST['rcpnmesrch'];

        if($_SERVER["REQUEST_METHOD"] == "POST") {
            $sql = "SELECT * FROM recipe_table WHERE recipe_name = '".$postrcpnmesrch."'";
            $result = mysqli_query($db, $sql);
            $row = mysqli_fetch_assoc($result);
        }
    }
?>
<html>
    <head>
        <title>Fresh Veg Admin, Edit Recipes</title>
        <link rel=stylesheet type="text/css" href="/FVRC/FVRC.css">
    </head>
    <body>
        <h1>Fresh Veg Admin, Edit Recipes</h1>
        <p>User:<b> <?PHP echo $login_session; ?></b>, logged in.</p>
        <form action="/FVRC/addeditrecipes.php">
            <button type="submit">Back</button>
        </form>
        <br/>
        <hr/>
        <table>
            <form method="POST">
                <tr>
                    <td>
                        <input type="text" placeholder="Search for recipes" name="rcpnmesrch">
                    </td>
                    <td>
                        <button type="submit">Search</button>
                    </td>
                </tr>
            </form>
        </table>
        <br/>
        <hr/>
        <form action="editrecipecheck.php" method="POST">
            <table cellspacing=0 cellpadding=5>
                <tr>
                    <td>
                    </td>
                    <td>
                        <label>Recipe Name:</label>
                    </td>
                    <td>
                        <input type="text" placeholder="Recipe Name" name="rcpnme" value="<?php echo $row['recipe_name']; ?>" required>
                    </td>
                </tr>
                <tr>
                    <td>
                        <label>Ingredient One:</label>
                    </td>
                    <td>
                        <input type="text" placeholder="Ingredient" name="ingnme1" required>
                    </td>


PHP中的变量范围是按函数定义的,变量不是
if
的本地变量。
var\u dump($row)
显示了什么?我很清楚,您有两个表单;其中一个不是用
关闭的,除非您忽略了它。另外,
不能成为
的子项。我也没有看到带有
提交
名称属性的提交按钮;当用户第一次打开页面时,没有提交表单,它不会进行搜索,并且不会分配
$row
。顺便说一句,该错误来自您的输入。最多使用三元运算符。
<td>
    <input type="text" placeholder="Recipe Name" name="rcpnme" value="<?php echo $row['recipe_name']; ?>" required>
</td>
<br />
<b>Notice</b>:  Undefined index: recipe_name in <b>E:\Program Files\xampp\htdocs\FVRC\editrecipe.php</b> on line <b>52</b>
<br />