Harmiih真的帮我用我的脚本(http://stackoverflow.com/questions/7510546/html-form-name-php-variable)。基本上我有一张有问题的桌子。表单从表格中检索问题,然后我需要选择的答案转到answer.php。一切工作与无线电按钮,但与复选框它只发送最后选中的复选框。有人可以帮帮我吗?形式与单选按钮和复选框发送到PHP
形式
$sql1="SELECT * FROM ex_question WHERE test_name = '$tid' AND q_type = 'mr' ORDER BY RAND() LIMIT 5";
$result1=mysql_query($sql1);
echo "<form method='post' action='answer.php'>";
while($row1 = mysql_fetch_array($result1))
{
$test_name=$row1['test_name'];
$q_nr=$row1['q_nr'];
$q_type=$row1['q_type'];
$question=$row1['question'];
$option1=$row1['option1'];
$option2=$row1['option2'];
echo "<P><strong>$q_nr $question</strong><BR>";
if ($q_type != 'mr') {
if($option1!="") {
echo "<input type='radio' name='question[$q_nr]' value='$option1'>$option1<BR>";
} else {
echo '';
}
if($option2!="") {
echo "<input type='radio' name='question[$q_nr]' value='$option2'>$option2<BR>";
} else {
echo '';
}
} else {
if($option1!="") {
echo "<input type='checkbox' name='question[$q_nr]' value='$option1'>$option1<BR>";
} else {
echo '';
}
if($option2!="") {
echo "<input type='checkbox' name='question[$q_nr]' value='$option2'>$option2<BR>";
} else {
echo '';
}
}
echo "<BR>";
echo "<BR>";
echo "</p>";
}
echo "<input type='submit' value='Send Form'>";
echo "</form>";
answer.php
<?php
//Key is $q_nr and $answer is selected $option
foreach($_POST['question'] as $key => $answer) {
echo $key;
echo $answer;
}
?>
谢谢JNDPNT的回答帮我,在我的形式我用: 名称= '的问题[$ q_nr] []' 和我answer.php: 的foreach($ _ POST [ '问题'] as $ key => $ answer){ echo $ answer [0]; echo $ answer [1]; } – Wilest
顺便说一下,你不必手动回复所有的复选框答案,因为你不知道用户将选择多少复选框。一个例子,你可以做到这一点。 [链接](http://pastebin.com/vjS44ZMi) – Harmiih
@Harmiih谢谢,现在就试试吧。 – Wilest