如何在 php 中将多行作为 json 传递

Question

我正在尝试将行详细信息作为 JSON 传递。如何将所选列与查询分开

$SQLTask = "
SELECT priority
     , task_name AS title
     , task_description AS description
     , start_date AS start
     , end_date AS end 
  FROM t_tasks
"; -- removed extraneous semi-colon
    $RESULT_Task = mysqli_query( $conn, $SQLTask );
            
    while($ROW_TASK = mysqli_fetch_assoc($RESULT_Task)) 
    {
        $TASK_NAME = $ROW_TASK['title'];

        $TASK_DESC = $ROW_TASK['description'];

        $TASK_START = $ROW_TASK['start'];

        $TASK_END = $ROW_TASK['end'];

    }
   

        $TASK_DETAILS[] = array('title' => $TASK_NAME, 'description' => $TASK_DESC, 'start' => $TASK_START, 'end' => $TASK_END); 

mysqli_close($conn);

echo json_encode($TASK_DETAILS);

按下面的格式试过它并且有效，但我需要将优先级列用于其他查询。 JSON 用于提供日历并显示事件。

<?php 
require('../classes/class_connect.php');

$Project_ID = $_GET['id'];

    $SQLTask = "SELECT task_name AS title, task_description AS description, start_date AS start, end_date AS end FROM t_tasks WHERE project_id = " . $Project_ID . "";
    $RESULT_Task = mysqli_query( $conn, $SQLTask );
            
    while($ROW_TASK = mysqli_fetch_assoc($RESULT_Task)) 
    {
        $TASK_ARRAY[] = $ROW_TASK;

        // if($ROW_TASK['priority'] == 'Low')
        // {
        //     $eventClassNames = "bg-light-secondary";
        // }
        // else if($ROW_TASK['priority'] == 'Normal')
        // {
        //     $eventClassNames = "bg-light-primary";
        // }
        // else if($ROW_TASK['priority'] == 'High')
        // {
        //     $eventClassNames = "bg-light-warning";
        // }
        // else if($ROW_TASK['priority'] == 'Urgent')
        // {
        //     $eventClassNames = "bg-light-danger";
        // }

        // array_push($TASK_ARRAY,$eventClassNames);
    }
    mysqli_close($conn);

               
echo json_encode($TASK_ARRAY);```

Answer 1

<?php
$mysqli = new mysqli("localhost", "my_user", "my_password", "world");

/* check connection */
if ($mysqli->connect_errno) {
    printf("Connect failed: %s\n", $mysqli->connect_error);
    exit();
}

$query = "SELECT Name, CountryCode FROM City ORDER by ID LIMIT 3";
$result = $mysqli->query($query);

/* numeric array */
$row = $result->fetch_array(MYSQLI_NUM);
printf ("%s (%s)\n", $row[0], $row[1]);

/* associative array */
$row = $result->fetch_array(MYSQLI_ASSOC);
printf ("%s (%s)\n", $row["Name"], $row["CountryCode"]);

/* associative and numeric array */
$row = $result->fetch_array(MYSQLI_BOTH);
printf ("%s (%s)\n", $row[0], $row["CountryCode"]);

/* free result set */
$result->free();

/* close connection */
$mysqli->close();
?>

echo json_encode($row)

参考：https://www.php.net/manual/en/mysqli-result.fetch-array.php

Answer 2

对于“如何将所选列与查询分开”，我不确定我是否清楚地理解了您的问题以及您想要实现的目标。

但是对于您说明的第二部分，因为您想要的只是优先级字段，您可以从查询中删除所有其他字段。您也可以在查询中添加 eventClassNames。

尝试使用 mysqli 的 fetch_all 看看它是如何工作的。对于第一部分，您不必遍历结果集来创建数组，除非对数组进行进一步处理 https://www.php.net/manual/en/mysqli-result.fetch-all.php

$SQLTask = "SELECT priority,
                   case priority
                      when 'Low' then 'bg-light-secondary'
                      when 'Normal' then 'bg-light-primary'
                      when 'High' then 'bg-light-warning'
                      when 'Urgent' then 'bg-light-danger'
                   end AS eventClassName
            FROM t_tasks WHERE project_id = " . $Project_ID . "";
$RESULT_Task = mysqli_query( $conn, $SQLTask );
$TASK_ROWS=$RESULT_TASK->fetch_all(MYSQLI_ASSOC);

$TASK_ARRAY=array_column($TASK_ROWS,'eventClassName');

echo json_encode($TASK_ARRAY);

如何在 php 中将多行作为 json 传递

How to pass multiple rows as json in php

php

mysql

arrays

json

mysqli