当我选中复选框时，我需要一些js / ajax / jquery脚本动态地将数据保存到数据库。 当前的复选框或加载到记录旁边的复选框，并根据是否选中它来更改数据库中的变量。但是我必须在选择一个以将其保存到数据库后重新加载页面。我可以做其他一切，但了解如何实现ajax，所以我不必每次都提交查询和刷新页面。非常感谢任何帮助。
<form name="form1aa" method="post" action="process.php?fn=<? echo $rows['first']; ?>&class=<?php echo $rows['class']; ?>&last=<?php echo $rows['last']; ?>
&model=<?php echo $rows['model']; ?>&cas=<?php echo $rows['cases']; ?>&upid=<?php echo $id; ?>&group=1" id="form1a" >
<select name="type" onchange=" fill_damage (document.form1aa.type.selectedIndex); ">
<option value="Hardware">Hardware</option>
<option value="Software">Software</option>
</select>
<select name="damage">
</select>
<input type=text name="comment" placeholder="Comments Box">
<input type=text name="cost" placeholder="Cost">
<input type="submit" value="Save" name="Save">
</form>
<?php
//Job Status
if(isset($_POST['checkbox'])){$checkbox = $_POST['checkbox'];
if(isset($_POST['activate'])?$activate = $_POST["activate"]:$deactivate = $_POST["deactivate"])
$id = "('" . implode( "','", $checkbox ) . "');" ;
$sql="UPDATE repairs SET status = '".(isset($activate)?'Completed':'In Progress')."' WHERE id=$id" ;
$result = mysql_query($sql) or die(mysql_error());
}
//End Job Status
//Payment Status
if(isset($_POST['paycheck'])){$paycheck = $_POST['paycheck'];
if(isset($_POST['paid'])?$paid = $_POST["paid"]:$unpaid = $_POST["unpaid"])
$id = "('" . implode( "','", $paycheck ) . "');" ;
$sql="UPDATE repairs SET paid = '".(isset($paid)?'Paid':'Unpaid')."' WHERE id=$id" ;
$result = mysql_query($sql) or die(mysql_error());
}
//End Payment Status
//Return Status
if(isset($_POST['retcheck'])){$retcheck = $_POST['retcheck'];
if(isset($_POST['ret'])?$ret = $_POST["ret"]:$unret = $_POST["unret"])
$id = "('" . implode( "','", $retcheck ) . "');" ;
$sql="UPDATE repairs SET ret = '".(isset($ret)?'Retuned':'In Office')."' WHERE id=$id" ;
$result = mysql_query($sql) or die(mysql_error());
}
//End Return Status
$sql="SELECT * FROM $tbl_name";
if(isset($_POST['all'])){
$sql="SELECT * FROM $tbl_name";
}
if(isset($_POST['tpc'])){
$sql="select * from $tbl_name WHERE class LIKE '1%'";
}
if(isset($_POST['drc'])){
$sql="select * from $tbl_name WHERE class LIKE 'D%'";
}
if(isset($_POST['bsc'])){
$sql="select * from $tbl_name WHERE class LIKE 'B%'";
}
$result=mysql_query($sql);
?>
<form name="frmactive" method="post" action="">
<input name="activate" type="submit" id="activate" value="Complete Job" />
<input name="paid" type="submit" id="Payment" value="Payment Status" />
<input name="ret" type="submit" id="ret" value="Returned 2 Student" />
<br />
<a id="displayText" href="javascript:toggle();">Show Extra</a>
<div id="toggleText" style="display: none">
<br />
<input name="unret" type="submit" id="unret" value="In Office" />
<input name="unpaid" type="submit" id="unpaid" value="Not Paid" />
<input name="deactivate" type="submit" id="deactivate" value="In Progress" /></div>
<table width="1000" border="0" cellpadding="3" cellspacing="1">
<thead>
<th width="67" align="center"><strong>Start Date</strong></th>
<th width="50" align="center"><strong>Cases</strong></th>
<th width="34" align="center"><strong>Type</strong></th>
<th width="120" align="center"><strong>Damage</strong></th>
<th width="31" align="center"><strong>Comment</strong></th>
<th width="31" align="center"><strong>Cost</strong></th>
<th width="90" align="center"><strong>Payment Status</strong></th>
<th width="100" align="center"><strong>Returned 2 Student</strong></th>
<th width="100" align="center"><strong>Job Status</strong></th>
</thead>
<?php
while($rows=mysql_fetch_array($result)){
?>
<tr>
<td><? echo $rows['start']; ?></td>
<td><? echo $rows['cases']; ?></td>
<td><? echo $rows['type']; ?></td>
<td width="70"><? echo $rows['damage']; ?></td>
<td width="70"><? echo $rows['comment']; ?></td>
<td><? echo "$"; echo $rows['cost']; ?></td>
<!--Payment Display(Start)-->
<?php
if($rows['paid']=="Paid")
{
?>
<td><input name="paycheck[]" type="checkbox" id="paycheck[]" value="<? echo $rows['id']; ?>">
<? echo $rows['paid'];?>
</td>
<?
}
if($rows['paid']=="Unpaid")
{
?>
<td width="21"><input name="paycheck[]" type="checkbox" id="paycheck[]" value="<? echo $rows['id']; ?>">
<? echo $rows['paid']; ?>
</td>
<?
}
if($rows['ret']==""){
?>
<td width="50">No Data</td>
<?
}
?>
答案 0 :(得分：3)
使用jQuery做，一个简单的例子可能是：
<强> HTML：
<input type="checkbox" name="option1" value="Milk">
<input type="checkbox" name="option2" value="Sugar">
<input type="checkbox" name="option3" value="Chocolate">
<强> JS：
$("input[type='checkbox']").on('click', function(){
var checked = $(this).attr('checked');
if(checked){
var value = $(this).val();
$.post('file.php', { value:value }, function(data){
// data = 0 - means that there was an error
// data = 1 - means that everything is ok
if(data == 1){
// Do something or do nothing :-)
alert('Data was saved in db!');
}
});
}
});
PHP：file.php
<?php
if ($_POST && isset($_POST['value'])) {
// db connection
$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!$link) {
// error happened
print(0);
}
mysql_select_db('mydb');
// sanitize the value
$value = mysql_real_escape_string($_POST['value']);
// start the query
$sql = "INSERT INTO table (value) VALUES ('$value')";
// check if the query was executed
if(mysql_query($sql, $link)){
// everything is Ok, the data was inserted
print(1);
} else {
// error happened
print(0);
}
}
?>
答案 1 :(得分：1)
简单地说......
$('input:checkbox').click( function() {
clicked = $(this).attr('checked');
if (clicked) {
/* AJAX the server to tell them it was clicked. */ }
else {
/* AJAX the server to tell them it was unclicked. */ } } );
答案 2 :(得分：0)
我可以使这更简单。首先，你需要打一个复选框!!
<form name="form1aa" method="post" action="process.php?fn=<? echo $rows['frist']; ?>&class=<?php echo $rows['class']; ?>&last=<?php echo $rows['last']; ?>
&model=<?php echo $rows['model']; ?>&cas=<?php echo $rows['cases']; ?>&upid=<?php echo $id; ?>&group=1" id="form1a" >
<select name="type" onchange="fill_damage(document.form1aa.type.selectedIndex);">
<option value="Hardware">Hardware</option>
<option value="Software">Software</option>
</select>
<select name="damage">
</select>
<input type="text" name="comment" placeholder="Comments Box">
<input type="text" name="cost" placeholder="Cost">
<input type="checkbox" name="somecheck" onchange="if(this.checked)document.form1aa.submit()">Check this to Save.
<input type="submit" value="Save" name="Save">
</form>
<script type="javascript>
//another function that works for onchange="dosubmit(this)"
//IF you put it after the form.
function dosubmit(el) {
if (el.checked) {
document.form1aa.submit();
}
}
</script>
尽可能删除onchange事件中的空格。
答案 3 :(得分：0)
如果您有动态复选框列表，并且想要将点击的动态保存到数据库或插入未选中的复选框，请按以下步骤操作：
HTML/PHP
<?php
// $checklists are models that I am getting from db
$checklists = CheckList::getCheckLists(3);
echo '<ul>';
foreach ($checklists as $checklist) {
$isChecked = $checklist->getAnswer($requestID, $checklist->primaryKey);
$checked = $isChecked ? "checked" : "";
echo '<li>';
echo "<input id='{$checklist->primaryKey}'
name='{$checklist->primaryKey}' type='checkbox' {$checked}
value='{$isChecked}' data-request-id='{$requestID}'>
$checklist->check_list_text";
echo '</li>';
}
echo '</ul>';
?>
Jquery
<script>
$("input[type='checkbox']").on('click', function(){
var checkbox = $(this);
var checked = checkbox.prop('checked');
var checklistId = checkbox.attr("id");
$.ajax({
url:"<?= Url::to(['default/add-checklist-answer']) ?>",
// I don't need to write the type here because I am using Yii Framework
// type: 'post',
data: {
checklistId: checklistId,
requestId: checkbox.data('request-id'),
checked: checked
},
success: function(data) {
//alert(data);
console.log(data.firstMessage)
},
error: function(data) {
// alert(data);
}
});
});
</script>
我希望它适用于 MVC 用户。