php - 属性内的 If else 语句和 If else 语句内的属性

Question

我正在处理一个标签，我可以在其中拥有一个具有数据映射突出显示的链接图像，我如何才能使 if else 语句在属性中起作用？并在声明中包含另一个属性。

这是我的代码:

<map id="ground" name="ground">
    <?php

    $sql = "SELECT * FROM stall s 
    LEFT JOIN tenant t 
    ON t.stall_id = s.stall_id
    LEFT JOIN rent r
    ON r.tenant_id = t.tenant_id 
    AND r.rent_status = 1
    WHERE s.stall_id = 1";          

    $query = $conn->prepare($sql);
    $query->execute();
    $fetch = $query->fetchAll();

    foreach ($fetch as $key => $value) { ?>
    <area id="9" shape="rect" coords="9,98,46,117" data-toggle="modal" class = "stallstyle1"  
    href="#stall_modal9" 
    value="<?php echo $value['rent_status']?>"
    <?php
    if ($value ['rent_status'] == 1) {
    data-maphilight='{"strokeColor":"008000","strokeWidth":3,"fillColor":"7CFC00","fillOpacity":0.6,"alwaysOn":true}'>
     } else {
    data-maphilight='{"strokeColor":"FF0000","strokeWidth":3,"fillColor":"FF3333","fillOpacity":0.6,"alwaysOn":true}'>
    }
    ?>

<?php } ?> 

Answer 1

使用三元运算符比使用 if/else 更容易理解。因此，使用三元运算符获取属性 data-maphilight 的值。然后在您的标签中使用它。如果您从数据库中获取值，希望以下代码能够正常工作。

    <map id="ground" name="ground">
    <?php

    $sql = "SELECT * FROM stall s 
    LEFT JOIN tenant t 
    ON t.stall_id = s.stall_id
    LEFT JOIN rent r
    ON r.tenant_id = t.tenant_id 
    AND r.rent_status = 1
    WHERE s.stall_id = 1";          

    $query = $conn->prepare($sql);
    $query->execute();
    $fetch = $query->fetchAll();

    foreach ($fetch as $key => $value){

    $maphilight= ($value['rent_status'] == 1)?'{"strokeColor":"008000","strokeWidth":3,"fillColor":"7CFC00","fillOpacity":0.6,"alwaysOn":true}':'{"strokeColor":"FF0000","strokeWidth":3,"fillColor":"FF3333","fillOpacity":0.6,"alwaysOn":true}';
     ?>
    <area id="9" shape="rect" coords="9,98,46,117" data-toggle="modal" class = "stallstyle1"  
    href="#stall_modal9" value="<?php echo $value['rent_status'];?>" data-maphilight='<?php echo $maphilight;?>'>
<?php } ?>