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我有分頁的表格和輸入字段。我想讓用戶創建新的流程。創建表允許用戶查看已經存在的進程。 HMTL代碼：沒有帶分頁表的彈出窗口
<div class="inner">
<center><table class="paginated">
<?php
$result = getProcessData();
if (mysqli_num_rows($result) > 0)
{
echo '
<thead>
<tr>
<th>Process</th>
<th>Created By</th>
</tr>
</thead>';
}
$count = 0;
if (mysqli_num_rows($result) > 0)
while($row = mysqli_fetch_array($result))
{
$process = $row['theme_name'];
$createdBy = $row['createby'];
echo '
<tbody>
<tr valign="top">
<td>' .$process. '</td>
<td>' .$createdBy. '</td>
</tr>
</tbody>';
$count++;
}
?>
</table></center>
<form id="reguserform" method="post" action="process.php#err" style="margin-top:40px;">
<?php
$res = verifyFormFields();
?>
<!-- Username field -->
<input type="text" class="input name" placeholder="Process name (required)" name="process" required/>
<!--<label class="tooll tool1">- No special Characters except _<br>- Character Limit: 4 to 20<br>- Username must be Unique</label>-->
<center><input type="submit" class="button small" name="submit" value="Create Process"/></center>
<div id="registerModal" class="reveal-modal small" data-reveal aria-labelledby="modalTitle" aria-hidden="true" role="dialog" data-options="close_on_background_click:false">
<h2 id="modalTitle">Success!</h2>
<div>
<div id="testRegister" style="font-weight: 400;font-size: 1.5em; font-family: 'Raleway', Arial, sans-serif;"></div>
<div class="right">
<a href="#" id="closebtn" onclick="popUpNo()" class="button">Ok</a>
</div>
</div>
</div>
<?php
if($count == 1)
{
if($res=="")
{
registerProcess();
echo"<script>document.getElementById('testRegister').innerHTML=registerStr;callShowAlert();</script>";
}
else{
echo "
<a style='color:red';>
$res
</a>
";
}
}
?>
</form>
</div>
我驗證的字段和插入進程到數據庫：
function registerProcess(){
$con = mysqli_connect(DATABASE_HOST, DATABASE_USER, DATABASE_PASSWORD, DATABASE_NAME);
global $process;
// To protect MySQL injection (more detail about MySQL injection)
$process = cleanInputData($process);
$process = mysqli_real_escape_string($con, $process);
//query result
$result = insertNewProcess(NULL,$process,$_SESSION['login_user'],1);
// Check result
if (!$result) {
$msg2= "Process is already added. You cannot add same process twice!";
echo $msg2;
die('Invalid queryyyy: ' . mysqli_error($con));
}
// the message
$msg = "Process is added!";
// use wordwrap() if lines are longer than 70 characters
echo "<script>var registerStr = 'Process Is Added';</script>";
//echo "<meta http-equiv='refresh' content='0'>";
}
function verifyFormFields(){
global $process;
if(empty($process)){
return "<p id ='err'>Please Input Process</p>";
}
else return "";
}
而且我使用JavaScript調用彈出窗口，並表分頁：
<script type="text/javascript">
(function() {
$(document).foundation();
});
function callShowAlert()
{
$('#registerModal').foundation('reveal', 'open');
}
function popUpNo()
{
$('#registerModal').foundation('reveal', 'close');
location.href = 'process.php';
}
var testVal = 0;
</script>
<script>
$(document).ready(function() {
$('table.paginated').each(function() {
var currentPage = 0;
var numPerPage = 5;
var $table = $(this);
$table.bind('repaginate', function() {
$table.find('tbody tr').hide().slice(currentPage * numPerPage, (currentPage + 1) * numPerPage).show();
});
$table.trigger('repaginate');
var numRows = $table.find('tbody tr').length;
var numPages = Math.ceil(numRows/numPerPage);
var $pager = $('<div class="pager"></div>');
for (var page = 0; page < numPages; page++) {
$('<span class="page-number"></span>').text(page + 1).bind('click', {
newPage: page
}, function(event) {
currentPage = event.data['newPage'];
$table.trigger('repaginate');
$(this).addClass('active').siblings().removeClass('active');
}).appendTo($pager).addClass('clickable');
}
$pager.insertBefore($table).find('span.page-number:first').addClass('active');
});
});
</script>
它工作完美，我彈出窗口和系統插入記錄到數據庫，但只有只有輸入字段和提交按鈕。當我把桌子，它不工作。系統不重新加載並且不將記錄插入到數據庫中。爲什麼？