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。

现在我有四个看起来像这样的按钮

用户单击上述按钮之一时，将显示此

因此，现在当用户（例如）选择construction并下一步选择（例如）Egypt' in the console and clicks button确认displays [855,599075], user can select multiple countries, this works as expected for构造,功率,油时，

现在我想要的是，如果用户例如在这四个按钮中单击All available industries按钮，然后再选择Egypt并单击confirm，它应该显示埃及建筑，石油，电力部门的项目总数之和855+337+406 = 1598，两个部门的预算总额之和1136173

这是我的解决方法

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;
              }
        });
    }
});

这里是php，用于获取所有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 industries btn并选择在控制台上获得[0,0]的国家/地区时。

我需要更改以获得我想要的东西吗？任何帮助或建议，将不胜感激，

Answer 1

您可以使用SQL JOIN运算符，否则在这种情况下，隐式联接将是最干净的：

$result = mysqli_query($conn, "SELECT * FROM construction, power, oil_and_gas, industrial WHERE construction.Countries = power.Countries AND power.Countries = oil_and_gas.Countries AND oil_and_gas.Countries = industrial.Countries");

您需要WHERE条件，以便它知道每个不同表的行如何相互关联。您可以使用表的别名将其缩短一点：

$result = mysqli_query($conn, "SELECT * FROM construction as C, power as P, oil_and_gas as G, industrial as I WHERE C.Countries = P.Countries AND P.Countries = G.Countries AND G.Countries = I.Countries");

但是，在这种情况下，我认为您可能需要考虑更改数据库的结构。好像您在它们之间重复了很多列。也许这些都可以放在一个表中，并带有一个“类型”列来指定它的功能，构造等。然后，您只查询一个表并按国家/地区名称分组即可获得所有结果，而不会造成凌乱的跨4个联接表。

Answer 2

您的dataAll.php

如果您选择了All available industries
您不检查扇区（最终您应该检查国家）因此您应该避免检查这种情况

<?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]);

Answer 3

单个表看起来不错。

（此答案的其余部分尚未完成，但可能很有用。）

首先，让我们设计将请求数据的URL。

.../foo.php?industry=...&country=...

但是，与其在客户端中特殊包装“全部”，不如在服务器中完成。也就是说，将生成用于工业的最后一个按钮

    ?industry=all

，PHP代码不会在WHERE子句中包含此代码：

    AND industry IN (...)

与&country=all和&country=egypt,iran,iraq相似

现在，让我简要介绍一下PHP：

$wheres = array();

$industry = @$_GET['industry'];
if (! isset($industry)) { ...issue error message or use some default... }
elseif ($industry != 'all') {
    $inds = array();
    foreach (explode(',', $industry) as $ind) {
        // .. should test validity here; left to user ...
        $inds[] = "'$ind'";
    }
    $wheres[] = "industry IN (" . implode(',', $inds) . )";
}

// ... repeat for country ...

$where_clause = '';
if (! empty($wheres)) {
    $where_clause = "WHERE " . implode(' AND ', $wheres);
}
// (Note that this is a generic way to build arbitrary WHEREs from the data)

// Build the SQL:

$sql = "SELECT ... FROM ... 
           $where_clause
           ORDER BY ...";
// then execute it via mysqli or pdo (NOT mysql_query)

现在，让我们谈谈使用AJAX。或不。有2个选择：

您本可以通过GET来调用PHP，并让PHP显示一个新页面。这意味着PHP将构建结果表。
您可能已经使用AJAX请求数据。这意味着Javascript将构造结果数据。

哪种选择可能取决于您更喜欢哪种语言。

使用php，ajax从数据库获取数据

3 个答案: