Get方法，然后将值传递到json格式 - Get method in php then pass the value into json format

Get method in php then pass the value into json format

本文关键字：json 格式值传方法然后 Get | 更新日期: 2023-09-27

我试图访问的值，我已经从get方法在我的PHP文件。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javascript文件中，我想使引用url与我已经在php

中的"get方法"中获得的url相同

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URL中，我希望URL与我的"Get方法"的值相同。例如:在我的get方法我有http://localhost:8888/ClicksChart/ckbox.php?CatId =1,2,3"。因此javascript文件中的url与PHP文件的值相同。有办法解决这个问题吗?谢谢您的帮助

您缺少一个'='和一个'echo'

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

你忘了写echo

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

您的参数在ajax中丢失

$.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');