以下是一个片段
<tr>
<td>7</td>
<td>Tick the Teaching Methods Used </td>
<td>
<input type="checkbox" id="lectures" name="lectures" value="lectures">Lectures
<input type="checkbox" id="study" name="study" value="study">Case Study
<input type="checkbox" id="audvid" name="audvid" value="audvid">Audio|Video
<input type="checkbox" id="interactive" name="interactive" value="interactive">Interactive Methods
<input type="checkbox" id="discussion" name="discussion" value="discussion">Discussion
<input type="checkbox" id="role" name="role" value="role">Role Play
<input type="checkbox" id="quiz" name="quiz" value="quiz">Quiz
</td>
</tr>
,验证码为
if ((document.form1.lectures.checked == false)
&& (document.form1.study.checked == false)
&& (document.form1.audvid.checked == false)
&& (document.form1.interactive.checked == false)
&& (document.form1.discussion.checked == false)
&& (document.form1.role.checked == false)
&& (document.form1.quiz.checked == false)) {
alert("Please check any one method");
isValid = false;
}
return isValid;
如何仅将选中的值插入mysql数据库,内爆似乎没有帮助
修改:如果我对所有复选框内幕使用相同的“名称”,但在这种情况下,我无法验证
答案 0 :(得分:0)
对所有复选框使用相同的名称。你说implode
,这很好。至于验证,将其更改为使用ID:
else if ((document.getElementById("lectures").checked == false)
&& (document.getElementById("study").checked == false)
&& (document.getElementById("audvid").checked == false)
&& (document.getElementById("interactive").checked == false)
&& (document.getElementById("discussion").checked == false)
&& (document.getElementById("role").checked == false)
&& (document.getElementById("quiz").checked == false)) {
alert("Please check any one method");
isValid = false;
}
return isValid;
答案 1 :(得分:0)
如果选中(选中)复选框,则仅将其置于POST中。
PHP:
$keys = explode(',','lectures,study,audvid'); //get all keys
$result = array();
foreach ($keys as $key) {
if (isset($_POST[$key]) && $_POST[$key]) { //selected!
$result[] = $key;
}
}
print_r($result); //only selected checkeboxes are in result.