Sql小提箱爲您的方便here。一個更好的方法來從MySql數據生成這個JSON數組與PHP
我正在從MySql表中獲取數據並將其轉換爲json數組。一切運作良好,我有我想要的方式輸出,但有沒有辦法可以改進(縮短/改進)?數組鍵必須保持不變,並且需要匹配dbs輸出。
我的代碼
$stmt = $conn->prepare("select name as name, age as age, address as address, pincode as pincode from json where name = 'peter'");
$stmt->execute();
while($row = $stmt->fetch()) {
#echo $row['name']." ".$row['age']." ".$row['address']." ".$row['pincode'].'<br>';
$myarray['name'] = $row['name'];
$myarray['age'] = $row['age'];
$myarray['address'] = $row['address'];
$myarray['pincode'] = $row['pincode'];
}
echo json_encode($myarray);
我的輸出(上述代碼成功地輸出該)
{"name":"Peter","age":"30","address":"1 Elm Street","pincode":"91550"}
名稱是唯一的嗎?爲什麼你做「名稱爲名稱」,因爲它不是必需的 – Haagenti
名稱是唯一的。我將使用'where name ='myName'' – jmenezes