動態更新的複選框

我有一個複選框，可以在使用PHP和Ajax檢查/取消選中時動態更新MySQL數據庫。動態更新的複選框

我正在嘗試傳遞用戶名，以便Ajax腳本可以用用戶全名更新數據庫。

我把名字保存在名爲$ full_name的變量中。我似乎無法得到這個工作，雖然。請參閱下面的代碼：

的Javascript：

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script> 
<script> 
function chkit(uid, chk) { 
    chk=document.getElementById("chk").checked; 

    $.ajax({ 
     type: 'GET', 
     url: 'ajax.php', 
     data: { chkYesNo: chk, record_id: uid, full_name: user}, 
     success:function(data){ 
     // successful request; do something with the div background color 
     if(data==1) 
     { 
      $("#replace").addClass("checked_div_status").removeClass("replace");//removing first class and adding second class 
     } 
     else 
     { 
      $("#replace").addClass("replace").removeClass("checked_div_status");//removing second class and adding first class 
     } 
     } 
    }); 
} 
</script>

HTML：

<?php 

$record_id = $_GET['veh_id']; 
include '../dbconnect.php'; 
//fetching data from database 
$select=mysql_fetch_array(mysql_query("select invoice_checked from vehicle_details where veh_id = '$record_id' ")); 
?> 

<!--The checkbox whose enable to change div his background color and onclick call function to update database--> 

<table width=「100%」> 

<td id="replace2" class="<?php if($select['invoice_checked']==1) { echo 'checked_div_status2'; } else{ echo 'replace2'; } ?>"> 
<input name="chk2" type="checkbox" id="chk2" value="1" onclick="chkit2(<?php echo $record_id;?>,'chk2');" <?php if($select['invoice_checked']==1) { echo 'checked'; } else{ echo ''; } ?> /> 
&nbsp;Invoice Checked 
</td> 

</table>

Ajax.php：

<?php 
mysql_connect("server", "username", "password") or die("Could not connect: " . mysql_error()); 
mysql_select_db("database"); 
//here $get variable receive checkbox value true(1) either false(0) 
$get=$_GET['chkYesNo']; 
//here $get_id variable receive value of current id that you passed 
$get_id=$_GET['record_id']; 
$get_user=$_GET['full_name']; 
if($get=="true") 
{ 
    $mysql_query=mysql_query("update vehicle_details set hpi_registered='1', check_user='".$get_user."' where veh_id='".$get_id."'"); 
    $select=mysql_fetch_array(mysql_query("select hpi_registered from vehicle_details where veh_id='".$get_id."'")); 
    echo $select['hpi_registered']; 
} 
else 
{ 
    $mysql_query=mysql_query("update vehicle_details set hpi_registered='0', check_user='0' where veh_id='".$get_id."'"); 
    $select=mysql_fetch_array(mysql_query("select hpi_registered from vehicle_details where veh_id='".$get_id."'")); 
    echo $select['hpi_registered']; 
} 

?>

任何幫助將大大收到。

感謝，

約翰

來源

2014-11-05 John Higgins

某些調試lession你。請檢查我的意見：

// Do not need to replicate your code, if the same things happens in it. 
//instead, use a condition to set your variables, and use these variables later. 
if ($get == "true") { 
    $hpi_registered = 1; 
    //Escape your variable to avoid sql injection 
    $checkUser = mysqli_real_escape_string($conn, $_GET["full_name"]); 
} else { 
    $hpi_registered = 0; 
    $checkUser = 0; 
} 

//Store your query in a variable, so you can debug/dump it 
//Let's dump it, see, what is your query, and try to run in directly in sql. 
//Maybe it has syntax error. 
$sql = "UPDATE vehicle_details SET" 
    . " hpi_registered='" . intval($hpi_registered) . "'," 
    . " check_user='" . $checkUser . "'" 
    . " WHERE veh_id='" . intval($get_id) . "'"; 
mysqli_query($conn, $sql); 
//What happens, if you run it directly in sql? If this fails, now here is your 
//error. 

$sql = "SELECT hpi_registered" 
    . " FROM vehicle_details" 
    . " WHERE veh_id='" . intval($get_id) . "'"; 
//Do the same like previous query. 
$res = mysqli_query($conn, $sql); 
$select = mysqli_fetch_array($res); 

echo $select['hpi_registered'];

切勿使用mysql功能，因爲它們已被棄用。改爲使用mysqli或PDO。
通過轉義變量避免sql注入。

來源

2014-11-05 12:14:43 vaso123

動態更新的複選框

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