我收到错误“未定义的变量:第15行的C:\ xampp \ htdocs \ online test \ study_question1.php中的表”如何创建表并在其中添加数据。
<?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...";
}
}
?>
答案 0 :(得分:0)
您未在table
中声明study_question1.php
变量。你可以这样做:
Code1.php
<?php
$tableName = 'name';
header('Location: Code2.php?tableName=' . $tableName);
然后在Code2.php中
<?php
$tableName = $_GET['table'];
使用session也可以达到类似效果。
答案 1 :(得分:0)
在您的查询中,您使用$table
这是常量而不是仅使用table
而不使用$符号,您不能将常量用作变量。
希望这会对你有所帮助:)。