【发布时间】：2018-11-29 01:23:31
【问题描述】：
编辑：添加当前正在使用的数据
我创建了一个小的函数文件，作为数据库接口，用于以简单的方式处理查询。
// Run SQL query
function runSQLQuery($sql) {
// Open DB
$db = connectToDB();
if( $db->connect_errno ) {
$data['response'] = false;
$data['error_code'] = $db->connect_errno;
$data['error'] = $db->connect_error;
return $data;
}
// Run & return query
$response = $db->query($sql);
if( is_bool($response) ) {
$response = ($response) ? 'true' : 'false';
}
switch($response) {
case "true":
$data['raw'] = $response;
$data['response'] = true;
$data['error'] = null;
$data['id'] = $db->insert_id;
return $data;
case "false":
$data['raw'] = $response;
$data['response'] = false;
$data['error_code'] = $db->errno;
$data['error'] = $db->error;
return $data;
default:
$data['raw'] = $response;
$data['response'] = true;
$data['error'] = null;
if( $response->num_rows > 0) {
while( $row = $response->fetch_assoc() ) {
$data['data'][] = $row;
}
}
else {
$data['data'] = null;
}
return $data;
}
}
// Open database connection
function connectToDB() {
// Connect to DB
return new mysqli(
DB_SERVER,
DB_USER,
DB_PASS,
DB_NAME
);
}
在大多数情况下，我会在其他地方构建一个 SQL 查询并将其传递给这个 runSQLQuery 函数，然后将响应作为 JSON 回显给我的网络应用程序。
我对简单的 SELECT、UPDATE、DELETE 命令没有任何问题，都可以成功运行。但是我想为一个函数实现一些稍微复杂的东西，一个 SELECT INNER JOIN 查询。构建此查询并执行的函数是：
public function get($args) {
extract($args);
// Get event
if(isset($userId)) {
$sql =
"SELECT j.id, j.event_address as address, j.event_datetime as datetime, j.event_note as note, j.event_price as price, j.client_id as userId, j.client_card_id as cardId, GROUP_CONCAT(p.event_package_id) as packages
FROM nr_jobs as j
INNER JOIN nr_job_packages as p ON p.event_id = j.id
WHERE j.client_id = $userId
GROUP BY j.id;";
}
if(isset($jobId)) {
$sql =
"SELECT j.id, j.event_address as address, j.event_datetime as datetime, j.event_note as note, j.event_price as price, j.client_id as userId, j.client_card_id as cardId, GROUP_CONCAT(p.event_package_id) as packages
FROM nr_jobs as j
INNER JOIN nr_job_packages as p ON p.event_id = j.id
WHERE j.id = $jobId
GROUP BY j.id;";
}
$events = runSQLQuery($sql);
$events["sql"] = $sql;
return $events;
}
由于某种原因，运行此特定查询返回 false。我在我的网络应用程序上收到的 JSON 输出是
{
"raw": "false",
"response": false,
"error_code": 0,
"error": "",
"sql": "SELECT j.id, j.event_address as address, j.event_datetime as datetime, j.event_note as note, j.event_price as price, j.client_id as userId, j.client_card_id as cardId, GROUP_CONCAT(p.event_package_id) as packages FROM nr_jobs as j INNER JOIN nr_job_packages as p ON p.event_id = j.id WHERE j.client_id = 1 GROUP BY j.id;"
}
如果我直接运行此查询，我实际上会得到正确的结果，几行数据的包的 CONCAT 列为 1、2、3，正如预期的那样。
数据库结构：
CREATE TABLE IF NOT EXISTS nr_jobs(
id BIGINT NOT NULL AUTO_INCREMENT PRIMARY KEY,
event_address TEXT NOT NULL,
event_datetime DATETIME NOT NULL,
event_note TEXT,
event_price DECIMAL(13,4) NOT NULL,
client_id BIGINT NOT NULL,
client_card_id BIGINT NOT NULL,
FOREIGN KEY (client_id) REFERENCES nr_clients(id),
FOREIGN KEY (client_card_id) REFERENCES nr_payment_cards(id)
);
CREATE TABLE IF NOT EXISTS nr_job_packages(
id BIGINT NOT NULL AUTO_INCREMENT PRIMARY KEY,
event_package_id BIGINT NOT NULL,
event_id BIGINT NOT NULL,
FOREIGN KEY (event_id) REFERENCES nr_jobs(id) ON DELETE CASCADE,
FOREIGN KEY (event_package_id) REFERENCES nr_packages(id) ON DELETE CASCADE
);
查询的输出直接运行：
+----+-------------------------------------------- ------------------------------------------------+-------------- --------+------+------------+--------+--------+----- -----+ |编号 |地址 |日期时间 |注意 |价格 |用户名 |卡号 |包| +----+-------------------------------------------- ------------------------------------------------+-------------- --------+------+------------+--------+--------+----- -----+ | 1 | SIM HQ LT3.02A, 461 Clementi Rd, Singapore, South West 59, 新加坡 | 2018-11-28 10:38:00 | | 320.0080 | 1 | 3 | 2,3 | | 2 | Tay Eng Soon Library, Blk A, SIM HQ, Singapore, South West 59, 新加坡 | 2018-11-28 10:42:00 | | 320.0080 | 1 | 3 | 2,3,4 | | 3 | SIM HQ 公司办公室，461 Clementi Rd, Singapore, South West 59, Singapore | 2018-11-28 10:43:00 | | 1020.0000 | 1 | 3 | 1,2,3,4 | | 4 | L.A. City DOT, 洛杉矶, 加利福尼亚州 90012, 美国 | 2018-11-28 12:15:00 | | 860.0000 | 1 | 4 | 2,3,4 | +----+-------------------------------------------- ------------------------------------------------+-------------- --------+------+------------+--------+--------+----- -----+如果查询没有问题，并且其他查询在我的函数中工作，为什么这个特定的 SELECT INNER JOIN 查询失败且没有错误？
编辑：
nr_jobs 数据：
SELECT * FROM nr_jobs;
nr_job_packages
SELECT * FROM nr_job_packages;
【问题讨论】：
-
nr_job_packages 中有数据吗？
-
@MarcoPens 在编辑中添加。数据就在那里，并且可以通过直接查询正确检索。