Bil*_*lid 1 html javascript php codeigniter
我想验证我的表单.我在我的代码点火器视图中有一个表单验证javascript代码,但它无法正常工作,仍然在没有任何验证的情况下将值发送到下一页.你能告诉我我在哪里弄错了,为什么会这样?
码:
<form method="get" action="/calculator/stage_two" name="calculate" id="calculate" onsubmit="return validateForm();">
<div class="calc_instruction">
<input type="text" name="property_number" placeholder="No/Name" id = "property_number" class="stage_one_box" />
<input type="text" name="postcode" placeholder="Postcode" id = "postcode" class="stage_one_box" />
</div>
<input type = "image" name = "submit_calculator" id = "submit_calculator" value="Go" src = "/images/next_button.png" />
</form>
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Javascript功能:
<script type="text/javascript">
function validateForm() {
var postcode=document.forms["calculate"]["postcode"].value;
if (postcode==null || postcode=="") {
alert("Please enter the postcode to give you more accurate results");
document.forms["calculate"]["postcode"].focus();
return false;
}
</script>
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你错过了一个结束括号.验证代码中的任何错误都将返回非假值并允许提交
以下是使用表单访问的规范方法:
<form method="get" action="/calculator/stage_two" name="calculate"
id="calculate" onsubmit="return validateForm(this);">
<div class="calc_instruction">
<input type="text" name="property_number" placeholder="No/Name" id = "property_number" class="stage_one_box" />
<input type="text" name="postcode" placeholder="Postcode"
id = "postcode" class="stage_one_box" />
</div>
<input type = "image" name = "submit_calculator" id="submit_calculator" src="/images/next_button.png" />
</form>
<script type="text/javascript">
function validateForm(theForm) {
var postcode=theForm.postcode;
if (postcode.value=="") { // cannot be null or undefined if value="" on field
alert("Please enter the postcode to give you more accurate results");
postcode.focus();
return false;
}
return true; // allow submit
}
</script>
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