Question

我喜欢在yii2中识别查找中的第一个元素，但我还没有想出办法来做到这一点，

这是代码：

$services = TblWorksTags::find()->where(["active"=>true])->all();
foreach ($services as $service){
    echo '<li>'.$service->name.'</li>
}

从上面的代码我想让第一个项目具有不同的类，如

$services = TblWorksTags::find()->where(["active"=>true])->all();
foreach ($services as $service) {
    //if its the first element
    echo '<li class="active">'.$service->name.'</li>  //this has a diffrent  <li>

    //for the other elements
    echo '<li>'.$service->name.'</li>
}

Answer 1

您可以通过以下某些counter执行此操作： -

<?php
$services = TblWorksTags::find()->where(["active"=>true])->all();

$counter = 1;
foreach ($services as $service){
    if($counter ==1){
        //if its the first element
        echo '<li class="active">'.$service->name.'</li>';  // quote and ; missed in your post
    }else{
        //for the other elements
        echo '<li>'.$service->name.'</li>'; // quote and ; missed in your post

    }
$counter++;
} 
?>

我不知道Yii，所以如果以下代码： -

$services = TblWorksTags::find()->where(["active"=>true])->all();

为您提供索引数组（类似Array(0=>'something',1=>'something else', ......so on)）。那么您可以像下面这样使用它的索引： -

<?php
$services = TblWorksTags::find()->where(["active"=>true])->all();

foreach ($services as  $key=> $service){ //check $key is used here
    if($key == 0){
        //if its the first element
        echo '<li class="active">'.$service->name.'</li>';  // quote and ; missed in your post
    }else{
        //for the other elements
        echo '<li>'.$service->name.'</li>'; // quote and ; missed in your post
    }
} 
?>

使用yii2 findall并访问第一个元素

1 个答案: