我想創建一個借用按鈕的驗證,當按下但沒有項目沒有檢查它應該顯示提示,但發生了什麼是它不會顯示提示,也不會顯示在數據庫中。我將如何嘗試做到這一點。如何確定複選框是否已被檢查
以下是JavaScript代碼。 (我不知道這是否是正確的,即使)
$(".uniform_on").change(function(){
var max = 1;
if ($(".uniform_on:checked").length < max) {
$(".uniform_on").attr('disabled', 'disabled');
alert('Please Check the item of equipment to be borrowed!');
$(".uniform_on:checked").removeAttr('disabled');
} else {
$(".uniform_on").removeAttr('disabled');
}
})
請幫助我不知道大部分的代碼,因爲我只定製我自己的代碼,我對這種東西有限的知識。如果在未勾選複選框的情況下按下借用按鈕,該怎麼辦;我如何驗證這一點,即用戶在再次按借前先檢查某些東西。謝謝。
這是我借的整個代碼
<?php include('header.php'); ?>
<?php include('session.php'); ?>
<?php include('navbar_borrow.php'); ?>
<div class="container">
<div class="margin-top">
<div class="row">
<div class="alert alert-info">
<button type="button" class="close" data-dismiss="alert">×</button>
<strong><i class="icon-user icon-large"></i> Borrow Table</strong>
</div>
<div class="span12">
<form method="post" action="borrow_save.php">
<div class="span3">
<div class="control-group">
<label class="control-label" for="inputEmail">Borrower Name</label>
<div class="controls">
<select name="member_id" class="chzn-select"required/>
<option></option>
<?php $result = mysql_query("select * from member")or die(mysql_error());
while ($row=mysql_fetch_array($result)){ ?>
<option value="<?php echo $row['member_id']; ?>"><?php echo $row['firstname']." ".$row['lastname']; ?></option>
<?php } ?>
</select>
</div>
</div>
<div class="control-group">
<label class="control-label" for="inputEmail">Due Date</label>
<div class="controls">
<input type="text" class="w8em format-y-m-d highlight-days-67 range-low-today" name="due_date" id="sd" maxlength="10" style="border: 3px double #CCCCCC;" required/>
</div>
</div>
<div class="control-group">
<div class="controls">
<button name="delete_student" class="btn btn-success"><i class="icon-plus-sign icon-large"></i> Borrow</button>
</div>
</div>
</div>
<div class="span8">
<div class="alert alert-success"><strong>Select Equipment</strong></div>
<table cellpadding="0" cellspacing="0" border="0" class="table" id="example">
<thead>
<tr>
<th>Acc No.</th>
<th>Equipment Description</th>
<th>Category</th>
<th>Quantity</th>
<th>status</th>
<th>Add</th>
</tr>
</thead>
<tbody>
<?php $user_query=mysql_query("select * from equipment where status != 'Archive' ")or die(mysql_error());
while($row=mysql_fetch_array($user_query)){
$id=$row['equipment_id'];
$cat_id=$row['category_id'];
$cat_query = mysql_query("select * from category where category_id = '$cat_id'")or die(mysql_error());
$cat_row = mysql_fetch_array($cat_query);
?>
<tr class="del<?php echo $id ?>">
<td><?php echo $row['equipment_id']; ?></td>
<td><?php echo $row['equipment_description']; ?></td>
<td><?php echo $cat_row ['classname']; ?> </td>
<td><?php echo $row['quantity']; ?> </td>
<?php /* <td><?php echo $row['equipment_description']; ?></td> */ ?>
<td width=""><?php echo $row['status']; ?></td>
<?php include('tooltip_edit_delete.php'); ?>
<td width="20">
<input id="" class="uniform_on" name="selector[]" type="checkbox" value="<?php echo $id; ?>" >
</td>
</tr>
<?php } ?>
</tbody>
</table>
</form>
</div>
</div>
<script>
$(".uniform_on").change(function(){
var max = 1;
if($(".uniform_on:checked").length < max){
$(".uniform_on:checked").attr('disabled', 'disabled');
alert('Please Check the item of equipment to be borrowed!');
$(".uniform_on:checked").removeAttr('disabled');
}else{
$(".uniform_on").removeAttr('disabled');
}
})
</script>
</div>
</div>
</div>
<?php include('footer.php') ?>
做出改變和它驗證部分現在對於交易的一部分,我想借用它不會保存到數據庫工作但它只是提示你已經借 CODE新借按鈕:
<input name="delete_student" class="btn btn-success" type="button" value="Borrow" onClick="return validationfunction();" />
CODE爲的javascript:
<script>
function validationfunction() {
if($(".uniform_on:checked").length > 0) {
alert('Equipment has been borrowed.');
return true
}
else {
alert('Please check the item of equipment to be borrowed!');
return false;
}
}
</script>
這個我JavaScript和jQuery,而不是PHP ...另外,我看到0 ajax。 – SpYk3HH
向我們顯示你的'html' –