Question

我已经编写了代码，因此用户可以在第一个下拉列表中选择“客户名称”，然后选择“标记”并单击“提交”。

当用户点击“提交”时，他们将获得一个CSV文件作为输出，其值与所选的两个条件相匹配。

我无法将CSV作为输出。我认为在传递价值时我做错了什么。请帮忙。

主页上的代码是：

    <html>
<head>
</head><?php include "connection.php"; ?>
<script type="text/javascript" src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.1.0/jquery.js"></script>
<script>
function getTag(val)
{

    $.ajax({
        type: "POST",
        url: "get_tag.php",
        data: "ClientID="+val,
        success: function(data){

            $("#Tag_List").html(data);
        }

    });
}

function getCSV()
{
var e = document.getElementById("Client-list");
var strClient = e.options[e.selectedIndex].value;
var f = document.getElementById("Tag_List");
var strTag = f.options[f.selectedIndex].text;
$.ajax({
        type: "POST",
        url: "get_CSV.php",
        data: "ClientID="+strClient + "Tag="+strTag,
        success: function(){

            window.console.log('Successful');
            return false;
        }

    });
}
</script>
<body>
<form>
<label> Select Client Name </label>
<Select id="Client-list" onChange="getTag(this.value)">
<option value="">Select Client</option>
<?php 
$sql=" SELECT * FROM CLIENTDETAILS";
$result = $dbhandle->query($sql);
while ($rs=$result->fetch_assoc()){
?>
    <option value=" <?php echo $rs["ClientID"]; ?> "> <?php echo $rs["ClientName"] ?> </option>
    <?php
 }  ?>
</Select>
<label> Tag Name: </label>
<Select id="Tag_List">
<option value=""> Select Tag</option>
</Select>
<button value="submit" onClick="return getCSV();">Submit</button>
</form>
</body>
</head>
</html>

get_CSV.php上的代码是：

    <html>
<?php 

define("HOSTNAME","localhost");
define("USERNAME","root");
define("PASSWORD","");
define("DATABASE","mydb");

$con=new mysqli(HOSTNAME,USERNAME,PASSWORD,DATABASE) or die("Unable to connect");

$query="SELECT * FROM listdetails WHERE ClientID='".$_POST["ClientID"]."' AND Tag='".$_POST["Tag"]."';"

$result = mysqli_query($con, $query);
$row = mysqli_fetch_array($result, MYSQLI_ASSOC);

$fp = fopen('file.csv', 'w');

foreach ($row as $val) {
    fputcsv($fp, $val);
}

fclose($fp);


?>

</html>

无法将CSV作为OutPut获取 - PHP

0 个答案: