复选框数据动态保存到数据库点击

我需要一些js/ajax/jquery脚本将数据动态保存到数据库时，我检查复选框。当前的复选框或加载在记录旁边，并根据其选中或更改数据库中的变量。但我必须重新加载页面后，我选择一个将其保存到数据库。我可以做一切，但了解如何实现这个Ajax，所以我不必提交查询和每次刷新页面。任何帮助是极大的赞赏。复选框数据动态保存到数据库点击

<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> 
    <? 
} 
?>

来源

2012-05-02 Matthew

你试过的代码是什么？ – Satya

继承人我目前的代码，我只是没有能够让我的头与周围使用Ajax。 – Matthew

你将需要改变你的逻辑，检查我的答案有一个想法。 –

与jQuery做吧，一个简单的例子是：

HTML：

<input type="checkbox" name="option1" value="Milk"> 
<input type="checkbox" name="option2" value="Sugar"> 
<input type="checkbox" name="option3" value="Chocolate">

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); 
    } 
} 
?>

来源

2012-05-02 04:44:20

嘿我很抱歉，但我只有最小的经验与jquery你有手中的代码拉它在对此感到抱歉 – Matthew

我给你的代码做你需要的，开始去jquery.com，下载jquery。 js文件并将其导入到html中的脚本标记中。最后，把我给你的JS代码放在这个$（document）.ready（function（）{//在这里粘贴代码}）;这就是所有，你现在可以测试 –

我已经完成了所有这一切，代码不做任何事情，作为一个有害生物的遗憾。 – Matthew

简单放...

$('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. */ } });

来源

2012-05-02 04:33:37 DanRedux

我可以让这个更简单。首先，你需要添加一个复选框！

<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>

在可能的情况下摆脱onchange事件中的空间。

来源

2012-05-02 05:59:00

复选框数据动态保存到数据库点击

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