Question

我正在尝试访问我在php文件中获取get方法的值。我的PHP文件看起来像这样

<?php
include 'Con.php';
header('content-Type: application/json');

$catid = $_GET["CatId"];    

//array declaration
$array  = array();

//declaration for the index name of the array
$text1 = "data1";
$text2 = "data2";
$text3 = "data3";
$text4 = "data4";
$text5 = "data5";


$sql = "select `Total Cliks`,`Categories_idCategories`,`Month` from Clicks where Categories_idCategories in ($catid)";
$_sql = mysqli_query($connection,$sql);

foreach ($_sql as $result) {
    $Clicks = $result['Total Cliks'];
    $Categories_idCategories = $result['Categories_idCategories'];
    $Month = $result ['Month'];

    if(array_key_exists($Month, $array[$text1]) == false){
        $array[$text1][$Month] = $Clicks;   
    }
    elseif(array_key_exists($Month, $array[$text2]) == false){
        $array[$text2][$Month] = $Clicks;   
    }
    elseif(array_key_exists($Month, $array[$text3]) == false){
        $array[$text3][$Month] = $Clicks;   
    }
    elseif(array_key_exists($Month, $array[$text4]) == false){
        $array[$text4][$Month] = $Clicks;   
    }
    elseif(array_key_exists($Month, $array[$text5]) == false){
        $array[$text5][$Month] = $Clicks;   
    }


}
echo json_encode($array);
?>

然后在我的Javascricpt文件中，我想使引用url与我已经在php中获取“get method”的url相同

所以javascript代码看起来像这样

$(document).ready(function(){
$.ajax({
    url : "http://localhost:8888/ClicksChart/ckbox.php?CatId <?php     $_GET["CatId"];?>",
    type : "GET",
    success : function(array){
        console.log(array);

        alert('Welcome');

在网址中，我希望网址与我的“获取方法”的值相同例如：在我的get方法中，我有http://localhost:8888/ClicksChart/ckbox.php?CatId = 1,2,3“。所以javascript文件中的url与PHP文件的值相同。有什么方法可以解决这个问题吗？谢谢你的帮助

Answer 1

ajax中缺少参数

$.ajax({
    url : "http://localhost:8888/ClicksChart/ckbox.php?CatId=<?php echo $_GET['CatId'];?>",
    type : "GET",
    dataType:'json',
    success : function(array){
        console.log(array);

        alert('Welcome');

Answer 2

你忘了写回声

{{1}}

Answer 3

您缺少'='和'echo'

$.ajax({
url : "http://localhost:8888/ClicksChart/ckbox.php?CatId=<?php echo $_GET["CatId"];?>",

在php中获取方法然后将值传递给json格式

3 个答案: