PHP创建表脚本并在表中插入数据

问题描述：

我收到一个错误“未定义的变量：表C：\ xampp \ htdocs \ online test \ study_question1.php on line 15”如何创建表和添加数据它。PHP创建表脚本并在表中插入数据

<?php 
    $connection = mysqli_connect('localhost','root','','userquestion') or die(mysqli_error($connection)); 
    define("table","table_name"); 
     if(isset($_POST['next'])){ 
      $table=$_POST['table_name']; 
      $query= "CREATE TABLE `userquestion`.`$table` (`id` INT(15) NOT NULL AUTO_INCREMENT , `question` TEXT CHARACTER SET utf8 COLLATE utf8_unicode_ci NOT NULL , `option1` TEXT CHARACTER SET utf8 COLLATE utf8_unicode_ci NOT NULL , `option2` TEXT CHARACTER SET utf8 COLLATE utf8_unicode_ci NOT NULL , `option3` TEXT CHARACTER SET utf8 COLLATE utf8_unicode_ci NOT NULL , `option4` TEXT CHARACTER SET utf8 COLLATE utf8_unicode_ci NOT NULL , `true_ans` TEXT CHARACTER SET utf8 COLLATE utf8_unicode_ci NOT NULL , PRIMARY KEY (`id`)) ENGINE = InnoDB;"; 
      $data = mysqli_query($connection,$query); 
      header('Location: study_question1.php'); 

      return $table; 
     } 
?> 

<?php 
    include "online.php"; 
    $connection = mysqli_connect('localhost','root','','userquestion') or die(mysqli_error($connection)); 
    if(isset($_POST['submit'])) { 
     $question = $_POST["question"]; 
     $option1 = $_POST['option1']; 
     $option2 = $_POST['option2']; 
     $option3 = $_POST['option3']; 
     $option4 = $_POST['option4']; 
     $true_ans = $_POST['true_ans']; 
     $query = "INSERT INTO `userquestion`.`$table` (question,option1,option2,option3,option4,true_ans) VALUES ('$question','$option1','$option2','$option3','$option4','$true_ans')"; 
     $data = mysqli_query($connection,$query); 
     if($data) { 
     echo "ab dusra dal..."; 
     } 
    } 

?>

你可以指定哪些代码对应'study_question1.php'？ – Ayush

其中包含online.php的第二个代码 –

答

你没有在study_question1.php声明table变量。你可以这样来做：

Code1.php

<?php 
$tableName = 'name'; 
header('Location: Code2.php?tableName=' . $tableName);

然后在Code2.php

<?php 
$tableName = $_GET['table'];

还可以实现使用会话类似的效果。

答

如在您的查询中，您使用$table这是常数，而不是使用table没有$ sign，您不能使用常量作为变量。

希望这会帮助你:)

PHP创建表脚本并在表中插入数据

相关推荐