[英]Passing radio buttons to be redirected to a new page
我正在尽力改写这个
我想要这样,以便当某人转到某个页面时,他们获得单选按钮的值,该单选按钮是一个链接,一旦单击“提交”,就将其发送到该页面。
这是我尝试过的代码
<form name = "quoted" method="get">
<input id = "poster" type="text" name="poster" required="required" placeholder = "Credited Individual."> <br>
<textarea class = "actual_quote" name = "actual_quote" required="required" placeholder = "Write the question here!"></textarea><br><br><br>
<div class = "checkboxes" required="required">
<h3 style = "margin-top:-20px;">Please select one catagory that the quote falls into.</h3>
<label for="x"><input type="radio" name="x" value="stupid.php" id = "x" checked="checked" /> <span>stupid</span></label><br>
<label for="x"><input type="radio" name="x" value="stupider.php" id = "x" /> <span>stupider</span> </label><br>
<label for="x"><input type="radio" name="x" value="stupidest.php" id = "x"/> <span>stupidest</span></label>
</div>
<input id = "submit1" type="submit"><br>
</form>
这是应该使其工作的代码,但是没有。
$(function(){
$('form').submit(function(event){
event.preventDefault();
window.location = $('input[type=radio]:checked').val();
});
});
我究竟做错了什么? 我只希望用户能够选择一个单选选项,并且一旦选择了该单选选项,然后按提交将其重定向到该值所指向的那个特定页面。
请帮忙! 任何建议欢迎! -Connor
尝试这个
<script type="text/javascript">
function get_action(form) {
form.action = document.querySelector('input[name = "x"]:checked').value;
}
</script>
<form name = "quoted" method="get" onsubmit="get_action(this);">
<input id = "poster" type="text" name="poster" required="required" placeholder = "Credited Individual."> <br>
<textarea class = "actual_quote" name = "actual_quote" required="required" placeholder = "Write the question here!"></textarea><br><br><br>
<div class = "checkboxes" required="required">
<h3 style = "margin-top:-20px;">Please select one catagory that the quote falls into.</h3>
<label for="x"><input type="radio" name="x" value="stupid.php" id = "x" checked="checked" /> <span>stupid</span></label><br>
<label for="x"><input type="radio" name="x" value="stupider.php" id = "x" /> <span>stupider</span> </label><br>
<label for="x"><input type="radio" name="x" value="stupidest.php" id = "x"/> <span>stupidest</span></label>
</div>
<input id = "submit1" type="submit"><br>
</form>
这将起作用。 如果可行,请投票。 :)
您的代码在jsfiddle上似乎运行良好。
$(function(){
$('form').submit(function(event){
event.preventDefault();
window.location = $('input[type=radio]:checked').val();
});
});
确保php文件有效并且可以正常工作。 如果页面正在加载但未显示,则可能是您的php文件有问题。 如果单击提交时没有页面加载,请尝试使用其他浏览器,因为此处的代码可以正常工作。
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