PHP：將兩個單獨的mysql查詢連接到同一個json數據對象中

Question

我正在嘗試將下面的mysql查詢結果網格化為單個json對象，但不太確定如何正確地執行它。

$id = $_POST['id'];

$sql = "SELECT contracts.po_number, contracts.start_date, contracts.end_date, contracts.description, contracts.taa_required, contracts.account_overdue, jobs.id AS jobs_id, jobs.job_number, companies.id AS companies_id, companies.name AS companies_name
FROM contracts
LEFT JOIN jobs ON contracts.job_id = jobs.id
LEFT JOIN companies ON contracts.company_id = companies.id
WHERE contracts.id = '$id'
ORDER BY contracts.end_date";

$sql2 = "SELECT types_id
FROM contracts_types
WHERE contracts_id = '$id'";

//return data
$sql_result = mysql_query($sql,$connection) or die ("Fail.");
$arr = array();
while($obj = mysql_fetch_object($sql_result)) { $arr[] = $obj; }
echo json_encode($arr); //return json

//plus the selected options
$sql_result2 = mysql_query($sql2,$connection) or die ("Fail.");
$arr2 = array();
while($obj2 = mysql_fetch_object($sql_result2)) { $arr2[] = $obj2; }
echo json_encode($arr2); //return json

這是當前的結果：

[{"po_number":"test","start_date":"1261116000","end_date":"1262239200","description":"test","taa_required":"0","account_overdue":"1","jobs_id":null,"job_number":null,"companies_id":"4","companies_name":"Primacore Inc."}][{"types_id":"37"},{"types_id":"4"}]

注意最后一節[{“types_id”：“37”}，{“types_id”：“4”}]如何放在root下的一個單獨的塊中。 我希望它嵌套在名為“types”之類的第一個分支中。

我認為我的問題更多地與Php數組操作有關，但我並不是最好的。

謝謝你的任何指導。

Answer 1

在輸出為JSON之前將結果組合到另一個結構中。 使用array_values將類型ID轉換為類型ID的數組。 另外，修復SQL注入漏洞。 使用PDO ，並假設錯誤模式設置為PDO :: ERRMODE_EXCEPTION：

$id = $_POST['id'];
try {
    $contractQuery = $db->prepare("SELECT contracts.po_number, contracts.start_date, contracts.end_date, contracts.description, contracts.taa_required, contracts.account_overdue, jobs.id AS jobs_id, jobs.job_number, companies.id AS companies_id, companies.name AS companies_name
        FROM contracts
        LEFT JOIN jobs ON contracts.job_id = jobs.id
        LEFT JOIN companies ON contracts.company_id = companies.id
        WHERE contracts.id = ?
        ORDER BY contracts.end_date");

    $typesQuery = $db->prepare("SELECT types_id
        FROM contracts_types
        WHERE contracts_id = ?");

    $contractQuery->execute(array($id));
    $typesQuery->execute(array($id));

    $result = array();
    $result['contracts'] = $contractQuery->fetchAll(PDO::FETCH_ASSOC);
    $result['types'] = array_values($typesQuery->fetchAll(PDO::FETCH_NUM));

    echo json_encode($result); //return json
} catch (PDOException $exc) {
    ...
}

如果$contractQuery最多返回一行，則將獲取行更改為：

    $result = $contractQuery->fetch(PDO::FETCH_ASSOC);
    $result['types'] = array_values($typesQuery->fetchAll(PDO::FETCH_NUM));

Answer 2

通過在SQL級別使用JOIN合並兩個查詢，您似乎可以獲得更好的服務。 但是，假設兩個數組的長度相等：

for ($x = 0, $c = count($arr); $x < $c; $x++) {
    if (isset($arr2[$x])) {
        $arr[$x] += $arr2[$x];
    }
}

echo json_encode($arr);

編輯：你需要從改變mysql_fetch_object到mysql_fetch_assoc使其正常工作。

Answer 3

你為什么使用2個不同的數組？ 我只想在$arr而不是$arr2添加第二個查詢的行。 這樣，您最終會得到一個包含2個查詢中所有行的數組。