Question

当我尝试在Jenkins作业中执行docker-compose up作为shell构建步骤命令时，我收到了响应docker-compose: command not found。有谁知道如何解决这个问题？ Jenkins本身就是在CentOS机器上托管的。

我已经添加了用户＆＃39; jenkins＆＃39;到了'sudoers＆＃39;和用户组：root，jenkins，docker。

Answer 1

将docker-compose可执行文件添加到PATH envvar。或者使用完全指定的路径，例如/usr/local/bin/docker-compose

Answer 2

你必须在Jenkins环境中安装docker compose。请按照以下链接安装https://docs.docker.com/compose/install/

Answer 3

我通过在Jenkins文件中设置环境PATH变量来解决此问题，如下所示：

public function package_recommender($restaurant_id, $count, $money)
{
    $response = array();
    $response['success'] = true;
    $response['message'] = "success";
    $connection = null;


    mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

    try {
        $connection = mysqli_connect("localhost", "root", "", PackageRecommender::DATABASE_NAME);

        $sqlQuery1 = "SELECT * FROM food where NOT foodCategory_id=6 And NOT foodCategory_id=7 And NOT foodCategory_id=8 And restaurant_id='$restaurant_id' And stock>='$count' And isActive = 1;";
        $result1 = $connection->query($sqlQuery1);

        $sqlQuery2 = "SELECT * FROM food WHERE foodCategory_id=7 And restaurant_id = '$restaurant_id' And stock>'$count' And isActive = 1 order by updateDate desc;";
        $result2 = $connection->query($sqlQuery2);


        $sqlQuery3 = "SELECT * FROM food WHERE foodCategory_id=8 And restaurant_id = '$restaurant_id' And isActive = 1 order by updateDate desc;";
        $result3 = $connection->query($sqlQuery3);

        $sqlQuery4 = "SELECT * FROM food WHERE foodCategory_id=6 And restaurant_id = '$restaurant_id' And isActive = 1 order by updateDate desc;";
        $result4 = $connection->query($sqlQuery4);

        $desertArray = $result2->fetch_assoc();
        $drinkArray = $result3->fetch_assoc();
        $mealArray = $result4->fetch_assoc();

        $packagesArray = array();

        if ($result1->num_rows > 0) {
            $foodArray = $result1->fetch_assoc();
            foreach ($foodArray as $key => $foodValue) {
                $package = array();
                $package['food'] = $foodValue;

                if (isset($mealArray[$key])) {
                    $package['meal'] = $mealArray[$key];
                }
                if (isset($drinkArray[$key])) {
                    $package['drink'] = $drinkArray[$key];
                }
                if (isset($desertArray[$key])) {
                    $package['desert'] = $desertArray[$key];
                }

                array_push($packagesArray, $package);
            }
            $response["package"] = $packagesArray;
        }

        echo json_encode($response);
    } catch (mysqli_sql_exception $e) {
        $response['success'] = false;
        $response['message'] = "error 101";
        die(json_encode($response));
    }
}

Answer 4

这是我所缺少的命令

for (int i = 1; i <= 10 ; i++)
    {
    int Random = (int)(Math.random()*100);
    this.txtOutput.setText("Password:"+ password + Random);
    }

我处死了，詹金斯很好

＆＃39;搬运工-撰写＆＃39;在Jenkins作业中执行命令时找不到错误

4 个答案: