new*_*bie 5 javascript forms validation input
我以表格模式制作了我的表格,如下所示:
<form name="register" method="post" action="#" onSubmit="return validasi()">
<table width="507" border="0">
<h1 class="title"><a href="#">Form Perubahan Password</a></h1>
<tr>
<td width="190" ><span id="usernameerror" class="style20">Masukkan Username </span></td>
<td width="319"><input name="username" type="text"></td>
</tr>
<tr>
<td><span id="passworderror" class="style20">Masukkan Password Lama</span></td>
<td><input name="pass" type="password"></td>
</tr>
<tr>
<td><span id="password1error" class="style20">Masukkan Password Baru</span></td>
<td><input name="pass1" type="password"></td>
</tr>
<tr>
<td><span id="password2error" class="style20">Ulangi Masukkan Password Baru</span></td>
<td><input name="pass2" type="password"></td>
</tr>
<tr>
<td> </td>
<td><input type="submit" name="Submit" value="Submit">
<input type="reset" name="reset" value="Reset"></td>
</tr>
</table>
</form>
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这是我用javascript验证的代码.看看这个..
<script language="javascript">
function checkName(register)
{
var eobj = document.getElementById('usernameerror');
var susername = register.username.value;
var oRE = /^[a-z0-9]+[_.-]?[a-z0-9]+$/i;
var error = false;
eobj.innerHTML = '';
if (susername == '') {
error = 'Error: Username tidak boleh kosong';
register.username.focus();
}
else if (!oRE.test(susername))
{
error="Salah format";
}
if (error)
{
register.username.focus();
eobj.innerHTML = error;
return false;
}
return true;
}
function validatePwd(register) /* old password verification */
{
var eobj = document.getElementById('passworderror');
var invalid = ' ';
var pw = register.pass.value;
var error = false;
eobj.innerHTML = '';
if (pw.length < 1)
{
error = 'Masukkan password anda';
}
else if (pw.indexOf(invalid) > -1)
{
error = 'Anda harus mengisi password';
}
if (error)
{
register.pass.focus();
eobj.innerHTML = error;
return false
}
return true;
}
function validatePwd1(register) /* password & retype-password verification */
{
var eobj1 = document.getElementById('password1error');
var eobj2 = document.getElementById('password2error');
var invalid = ' ';
var pw1 = register.pass1.value;
var pw2 = register.pass2.value;
var error = false;
eobj1.innerHTML = '';
eobj2.innerHTML = '';
if (pw1.length < 1)
{
error = 'Masukkan password anda';
}
else if (pw1.indexOf(invalid) > -1)
{
error = 'Anda harus mengisi password';
}
if (error)
{
register.pass1.focus();
eobj1.innerHTML = error;
return false
}
if (pw1 != pw2)
{
eobj2.innerHTML = ' password tidak sama, coba masukkan kembali password anda';
return false;
}
return true;
}
function validasi()
{
var form = document.forms['register'];
var ary = [checkName, validatePwd, validatePwd1];
var rtn = true;
var z0 = 0;
for (var z0 = 0; z0 < ary.length; z0++)
{
if (!ary[z0](form))
{
rtn = false;
}
}
return rtn;
}
</script>
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当我使用这个验证通常形成它的工作但在表模式,验证代码不起作用..帮助我解决这个问题... tq
您已在 form 中声明onSubmit="return validasi()",但您在其中定义了函数。为了进行验证工作,您需要为其定义函数。
function validasi(){
// validation code goes here
}
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