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感谢您调查我的问题。它与PHP POST / GET函数完全相关。这是我的代码：index.php Â ]]]]] ...instead of <?php echo $_POST["story"]; ?> I want to print 200+ words story relating to what User has chosen. 使用条件if()或switch语句来查看您的$ _POST ['story']是否已设置并等于您选择的故事之一。 //--> 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 $output = null; //--> Empty variable to hold display info from conditional //--> Now to see if the form element that selects story is set using isset if(isset($_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{ $output = //--> Set a default setting that displays output here if neither story is selected. } } 将变量$output回显到您希望在html文档中显示故事内容的位置 <div> <?=$output?>  <?php echo $output; ?> </div> 您需要从某个地方获得这个故事。到达那里的几种方法。如果要从文本文件加载： // 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!'; 如果要从数据库加载它： $conn = new mysqli($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!'; 这假设您有一个名为stories的表，其中包含字段story_id和story_text。您还可以有一个单独的文件，将所有故事分配到变量中（假设您确实只有几个，加载未使用的故事对性能的影响很小）： $stories['Seaside'] = <<<EOL Here is the seaside story. EOL; $stories['Mountain'] = <<<EOL Here is the mountain story. EOL; 然后在您的故事文件中，“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），简单： <?php echo $story_text; ?> 上面的选项中，如果您正在寻找简单易维护的内容，我会选择加载文本文件来获取您的故事文本。祝您讲故事好运。：）假设您要使用表单变量inside 您的故事。您需要使用标记{{ place }}之类的标记，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); 这将“一旦有{{名称}}探索{{地点}} ...”变成“有一次Светослав探索堪察加半岛...” >等如果我了解你，是真的。但这是不好的解决方案。您应该使用数据库。 <?php session_start(); $_SESSION['story'][] = $_POST["story"]; ?> <html> <body> <center> <h1>Welcome <?php echo $_POST["name"]; ?>, age of <?php echo $_POST["age"]; ?> to the wonderfull land of <?php echo $_POST["place"]; ?> <?php echo implode(' ', $_SESSION['story']); ?> <a href=index.php>Edit</a> </body> </html>

php post get

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如何在php中插入图像和日期

if（isset（$ _ POST ['post-submit']））{$ post_category_id = $ _POST ['post_category_id']; $ post_title = $ _POST ['post_title']; $ post_author = $ _POST ['post_author']; $ ...

php mysql sql post insert

回答 1 投票 0

我如何在php中插入图像和日期

if（isset（$ _ POST ['post-submit']））{$ post_category_id = $ _POST ['post_category_id']; $ post_title = $ _POST ['post_title']; $ post_author = $ _POST ['post_author']; $ ...

php mysql sql post insert

回答 1 投票 0

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