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php - 如果表单验证失败,请不要提交表单

转载 作者:行者123 更新时间:2023-11-30 13:04:33 26 4
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我写的

<form action='test.php' method='post'>
.............
<submit onclick='validate();'></submit>
</form>

function valiadte(){
if(true){
var form = getElementByID(........);
form.submit();
}
else alert('error');
}

但如果失败,它仍然会转到 test.php。如果失败,我如何避免它转到 test.php 或者我可以将一些消息传递给 test.php?

提前致谢。

原始源代码:

            <form id="tabS" method="post" action="stdRegister.php">
<label>Account</label>
<input id="stdAccount" name="account" type="text" value="" class="input-xlarge" onblur="validateAccountAndID(this,document.getElementById('stdAccountHelp'));">
<span class="help-block" id="stdAccountHelp"></span>
<label>Password</label>
<input id="stdPassword" name="password" type="password" value="" class="input-xlarge" onblur="validatePassword(this,document.getElementById('stdPasswordHelp'));">
<span class="help-block" id="stdPasswordHelp"></span>
<label>Name</label>
<input id="stdName" name="name" type="text" value="" class="input-xlarge" onblur="validateName(this,document.getElementById('stdNameHelp'));">
<span class="help-block" id="stdNameHelp"></span>
<label>IDNumber</label>
<input id="stdID" name="idNumber" type="text" value="" class="input-xlarge" onblur="validateAccountAndID(this,document.getElementById('stdIDHelp'));">
<span class="help-block" id="stdIDHelp"></span>
<label>Department</label>
<select id="stdDepartment" name="department">
<?php
while ($row = $result->fetch_assoc()){
echo"<option>";
echo $row['department'];
echo"</option>";
}
?>
</select>
<label>Grade</label>
<select id="stdGrade" name="grade">
<option>1</option>
<option>2</option>
<option>3</option>
<option>4</option>
</select>
<div>
<button class="btn btn-primary" onclick="submitStudetForm();">Create Account</button>
</div>
<span class="help-block" id="stdRegisterHelp"></span>
</form>

function submitStudentForm(){
var form = document.getElementById('tabP');
if(
validateAccountAndID(document.getElementById('stdAccount'),document.getElementById('stdAccountHelp'))
&& validatePassword(document.getElementById('stdPassword'),document.getElementById('stdPasswordHelp'))
&& validateName(document.getElementById('stdName'),document.getElementById('stdNameHelp'))
&& validateAccountAndID(document.getElementById('stdID'),document.getElementById('stdIDHelp')) ){
form.submit();
alert("success");
}
else {
alert("Something wrong in the form. Please check again!");
return false;
//window.location.href='index.php';
}
}

最佳答案

只需在 else 情况下使用函数 return false

function validate(){
if(true){
return true;
}
else {
alert('error');
return false;
}
}

然后您会将 html 更改为:

<input type="submit" value="Submit" onsubmit="validate()">

关于php - 如果表单验证失败,请不要提交表单,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16155931/

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