PHP：根据 POST / GET 用户输入加载内容

PHP：根据 POST / GET 用户输入加载内容

感谢您调查我的问题。它纯粹与 PHP POST / GET 函数有关。

这是我的代码：

索引.php

<html>

<body>

 <p>

<center>

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

<label for="name">Name:</label> <input type="text" name="name">

<label for="age">Age:</label><input type="number" size=2 name="age"><p>

<label for="place">Place:</label> <input type="text" name="place"><p>

<label for="storys">Choose a story:</label>

<select name="story" id="story">

<option value="sea">Sea side</option>

<option value="mount">Mountain</option>

</select>

<p>

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

<button type="reset" value="Reset">Reset</button>

</form>

</body>

</html>

故事获取.php

<html>

<body>

<center>

<h1>Welcome <font color=red><i><?php echo $_POST["name"]; ?></i></font>, age of <font color=red><i><?php echo $_POST["age"]; ?></i></font><br>

to the wonderfull land of <br><font color=red><i><?php echo $_POST["place"]; ?></i></font>

<?php echo $_POST["story"]; ?>

<p><a href=index.php>Edit</a>

</body>

</html>

我的代码工作正常，但<?php echo $_POST["story"]; ?>我不想打印与用户选择的内容相关的 200 多个单词的故事。我希望您理解我想要实现的目标，并能够提出简单的解决方案。提前致谢

天涯尽头无女友

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2回答

12345678_0001

你需要从某个地方得到那个故事。有几种方法可以到达那里。如果要从文本文件加载它：// Here we sanitize the story ID to avoid getting hacked:$story_id = preg_replace('#[^a-zA-Z0-9_ -]#', '', $_POST['story']);// Then we load the text file containing the story:$path = 'stories/' . $story_id . '.txt';$story_text = is_file($path) ? file_get_contents($path) : 'No such story!';如果你想从数据库加载它：$conn = new mysqli($servername, $username, $password, $dbname);$stmt = $conn->prepare("SELECT * FROM stories WHERE story_id = ? LIMIT 1");$stmt->bind_param("s", $_POST['story']);$stmt->execute();$result = $stmt->get_result();$story = $result->fetch_assoc();$story_text = !empty($story['story_text']) ? $story['story_text'] : 'No such story!';stories这假设您有一个名为字段story_id和的表story_text。你也可以有一个单独的文件，将你所有的故事分配到变量中（假设你真的只有几个，加载未使用的故事对性能的影响是最小的）：$stories['Seaside'] = <<<EOLHere is the seaside story.EOL;$stories['Mountain'] = <<<EOLHere is the mountain story.EOL;然后在你的故事文件中，你“加载”它：$story_text = !empty($stories[$_POST['story']) ? $stories[$_POST['story']] : 'No such story!';然后（使用上述任何选项），只需：<?php echo $story_text; ?>在上述选项中，如果您正在寻找简单且易于维护的东西，我会选择文本文件加载来获取您的故事文本。祝你讲故事好运。:)假设您想在故事中使用表单变量。您需要使用标记，例如{{ place }}，并在输出故事文本之前替换它们：$story_text = str_replace('{{ name }}', $_POST['name'], $story_text);$story_text = str_replace('{{ place }}', $_POST['place'], $story_text);这会将“曾几何时有 {{ name }} 探索 {{ place }}...”变成“曾几何时有 Светослав 探索堪察加半岛...”等。

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哔哔one

...instead of <?php echo $_POST["story"]; ?> I want to print 200+ words story relating to what User has chosen.使用条件if()或switch语句查看您的 $_POST['story'] 是否已设置并等于您的故事选项之一。//--> ON story_get.php //--> (Provided the $_POST variable in fact has the values assigned to the global array)//--> use var_dump($_POST) on story_get.php to check if the global $_POST array has //--> key/value pairs coming from your index.php page// set variables that have your story information. $seaside = //--> Your story about the sea side$mountain = //--> Your story about the mountain$output = null; //--> Empty variable to hold display info from conditional//--> Now to see if the form element that selects story is set using issetif(isset($_POST['story']){  //--> Now that we know the input in the form that holds the value for story isset  //--> Check to see if it is set and then declare a variable and assign it to that variable  $story = $_POST['story'];  if($story === 'sea'){    $output = $seaside;  }elseif($story === 'mount'){    $output = $mountain;  }else{    $output = //--> Set a default setting that displays output here if neither story is selected.  }}$output在 html 文档中回显您希望显示故事内容的变量<div>  <?=$output?>    <?php echo $output; ?></div>

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