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javascript - 使用javascript在表格模式下验证表单

转载 作者:搜寻专家 更新时间:2023-11-01 04:26:18 26 4
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我在表格模式下制作了表格 像这样:

<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>&nbsp;</td>
<td><input type="submit" name="Submit" value="Submit">
<input type="reset" name="reset" value="Reset"></td>
</tr>
</table>
</form>

这是我使用 javascript 的验证代码。检查一下..

<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>

当我在通常情况下使用此验证时,它的工作形式但是在表格模式下,验证代码不起作用..帮助我解决这个问题... tq

查看演示 http://jsfiddle.net/andricoga/u9eZz/

最佳答案

您已经在 form 中声明了 onSubmit="return validasi()" ,但是您在其中定义了函数。为了验证工作,您需要为此定义函数。

function validasi(){
// validation code goes here
}

关于javascript - 使用javascript在表格模式下验证表单,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/15469817/

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