我有三个 PHP 脚本。 main.php questions.php 和 values.php
这是代码
main.php
<html>
<head>
<title></title>
</head>
<body>
<h1>Be Prepare for the battle</h1>
<?php
$strTitle = "Begin";
$strLink = "<a href = 'question.php?ques_id=1'>" . $strTitle ."</a>";
echo $strLink;
?>
</body>
</html>
问题.php
<?php
require_once('../connect.php');
$quesSQL = mysql_query("SELECT * FROM `questions` WHERE `ques_id`=". $_GET["ques_id"]);
if(!mysql_num_rows($quesSQL) >= 1)
{
die('Complete.');
}
$next = $_GET["ques_id"];
while($row = mysql_fetch_array($quesSQL)) {
$id = $row['ques_id'];
$strTitle = $row['ques_title'];
echo "<li>" . $strTitle . "</li><br/>";
}
$optSQL = mysql_query("SELECT `options`,`values` FROM questions_options WHERE " . $id . "= ques_id");
echo "<form action=\"values.php\" method=\"POST\">";
while($row = mysql_fetch_array($optSQL) ) {
$strOptions = $row['options'];
$strValues = $row['values'];
echo "<input type =\"radio\" name =\"valueIn\" value=" . $strValues . " />" . $strOptions . "<br/>";
}
echo "</form>";
$strTitle = "<input type =\"submit\" value=\"Next\">";
$next = $next + 1;
$strLink = "<a href = 'values.php?ques_id=" . $next . "'>" . $strTitle ."</a>";
echo $strLink;
mysql_close();
?>
values.php
<?php
require_once('../connect.php');
$input = $_POST['valueIn'];
$ansSQL = mysql_query("SELECT `answer` FROM questions WHERE 1-".$_GET["ques_id"]."= ques_id");
$marks = 0;
if($input == $ansSQL)
{
$marks = $marks+1;
}
else
{
$marks = $marks+0;
}
echo $marks;
?>
现在的问题是我必须将一个值从第二个脚本 (questions.php) 传递到第三个脚本 (values.php)。 它来自单选按钮名称值“valueIn”中的
最佳答案
我不确定您为什么使用链接来处理表单中可能包含的内容。正如 anusha 所述，您应该像这样为 ques_id 使用隐藏的输入字段
问题.php
<?php
require_once('../connect.php');
$quesSQL = mysql_query("SELECT * FROM `questions` WHERE `ques_id`=". $_GET["ques_id"]);
if(!mysql_num_rows($quesSQL) >= 1)
{
die('Complete.');
}
$next = $_GET["ques_id"];
while($row = mysql_fetch_array($quesSQL)) {
$id = $row['ques_id'];
$strTitle = $row['ques_title'];
echo "<li>" . $strTitle . "</li><br/>";
}
$optSQL = mysql_query("SELECT `options`,`values` FROM questions_options WHERE " . $id . "= ques_id");
echo "<form action=\"values.php\" method=\"POST\">";
while($row = mysql_fetch_array($optSQL) ) {
$strOptions = $row['options'];
$strValues = $row['values'];
echo "<input type =\"radio\" name =\"valueIn\" value=" . $strValues . " />" . $strOptions . "<br/>";
}
$next = $next + 1;
$strLink = '<input type="hidden" name="ques_id" value="'.$next.'">';
echo $strLink;
$strTitle = "<input type =\"submit\" value=\"Next\">";
echo $strTitle;
echo "</form>";
mysql_close();
?>
然后在下一步中通过 $_POST 可以使用这两个变量，如下所示
$input = $_POST['valueIn'];
$ques_id = $_POST['ques_id'];
关于php - 将html属性值传递给php中的下一个脚本，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/13945837/