复选框数据在单击时动态保存到数据库

一尘不染

复选框数据在单击时动态保存到数据库

ajax

当我选中复选框时，我需要一些js / ajax /
jquery脚本将数据动态保存到数据库中。复选框或当前加载到记录旁边的复选框，并根据是否选中变量来更改数据库中的变量。但是我必须在选择一个页面以将其保存到数据库后重新加载页面。我可以做其他所有事情，但了解如何对此实现ajax，因此我不必每次都提交查询并刷新页面。任何帮助是极大的赞赏。

    <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>
    <?
}
?>

阅读 189

2020-07-26

共1个答案

一尘不染

使用jQuery来做，一个简单的例子可能是：

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);
    }
}
?>

2020-07-26