使用php，ajax从数据库获取数据

Question

我有一个简单的部分，其中显示数据库中的数据，我的数据库如下所示。

现在我有四个按钮，像这样

当用户单击以上按钮之一时，它将显示

所以现在，当用户如选择construction和未来选择，例如，Egypt' in the console and clicks button确认displays [855,599075], user can select multiple countries, this works as expected for建设,电力,oil`，

现在我想，如果用户点击如All available industries这四个按钮和下一个选择，例如按钮Egypt，点击confirm它应该显示在建筑，石油，电力行业的埃及项目总数的总和 855+337+406= 1598总预算的两个部门的总和1136173

这是我的解决方案

的HTML

<div id="interactive-layers">
    <div buttonid="43" class="video-btns">
        <span class="label">Construction</span></div>
    <div buttonid="44" class="video-btns">
        <span class="label">Power</span></div>
    <div buttonid="45" class="video-btns">
        <span class="label">Oil</span></div>
    <div buttonid="103" class="video-btns">
        <span class="label">All available industries</span>
    </div>
</div>

这是js ajax

$("#interactive-layers").on("click", ".video-btns", function(){
    if( $(e.target).find("span.label").html()=="Confirm" ) {

        var selectedCountries = [];

        $('.video-btns .selected').each(function () {
            selectedCountries.push( $(this).parent().find("span.label").html() ) ;
        });

        if( selectedCountries.length>0 ) {
            if(selectedCountries.indexOf("All available countries")>-1) {
                selectedCountries = [];
            }


        } else {

            return;
        }

        var ajaxurl = "";
        if(selectedCountries.length>0) {
            ajaxurl = "data.php";
        } else {
            ajaxurl = "dataall.php";

        }

        $.ajax({
            url: ajaxurl,
            type: 'POST',
            data: {
                    countries: selectedCountries.join(","),
                    sector: selectedSector
            },
            success: function(result){
                console.log(result);
                result = JSON.parse(result);
                $(".video-btns").each(function () {
                    var getBtn = $(this).attr('buttonid');
                    if (getBtn == 106) {
                        var totalProjects = $("<span class='totalprojects'>"+ result[0] + "</span>");
                        $(this).append(totalProjects)
                    }else if(getBtn ==107){
                        var resultBudget = result[1]
                        var totalBudgets = $("<span class='totalbudget'>"+ '&#36m' +" " + resultBudget +"</span>");
                        $(this).append( totalBudgets)
                    }
                });
                return;
              }
        });
    }
});

这是PHP来获取所有dataall.php

$selectedSectorByUser = $_POST['sector'];
 $conn = mysqli_connect("localhost", "root", "", "love");
 $result = mysqli_query($conn, "SELECT * FROM meed");
 $data = array();

 $wynik = [];
$totalProjects = 0;
$totalBudget = 0;

 while ($row = mysqli_fetch_array($result))
 {
    if($row['Sector']==$selectedSectorByUser ) {
     $totalProjects+= $row['SumofNoOfProjects'];
     $totalBudget+= $row['SumofTotalBudgetValue'];
    }
 }
 echo json_encode([ $totalProjects, $totalBudget ] );
exit();
?>

这是data.php

<?php

$selectedSectorByUser = $_POST['sector'];
$countries = explode(",", $_POST['countries']);

//var_dump($countries);
 $conn = mysqli_connect("localhost", "root", "", "meedadb");
 $result = mysqli_query($conn, "SELECT * FROM meed");
 $data = array();

 $wynik = [];
$totalProjects = 0;
$totalBudget = 0;

 while ($row = mysqli_fetch_array($result))
 {
    if($row['Sector']==$selectedSectorByUser && in_array($row['Countries'],$countries ) ) {
    // array_push($data, $row);
     $totalProjects+= $row['SumofNoOfProjects'];
     $totalBudget+= $row['SumofTotalBudgetValue'];
    }
 }

 // array_push($wynik, $row);
 echo json_encode([ $totalProjects, $totalBudget ] );
//echo json_encode($data);
exit();
?>

现在，当用户单击All available industriesbtn并选择一个国家/地区时，我将 [0,0]在控制台上进入。

我需要更改以获得我想要的东西？任何帮助或建议将不胜感激，

Answer 1

在你的dataAll.php中

如果选择All available industries
，则不检查扇区，因为您需要所有扇区（最终应检查国家），因此应避免对此情况进行检查

<?php

$conn = mysqli_connect("localhost", "root", "", "love");
$result = mysqli_query($conn, "SELECT * FROM meed");
$data = [];

$wynik = [];
$totalProjects = 0;
$totalBudget = 0;

while ($row = mysqli_fetch_array($result)) {
    $totalProjects += $row['SumofNoOfProjects'];
    $totalBudget += $row['SumofTotalBudgetValue'];
}
echo json_encode([$totalProjects, $totalBudget]);