此时我得到以下 JSON:
["1Sales & Consulting","2Pyments","3Investing","4Financing","5Cross Functional"]
但我想要一个合适的 JSON,例如:
[{"id":1, "name": "Sales & Consulting"}{"id": 2, "name": "Pyments"}{"id": 3, "Investing"}{"id": 4, "name": "Financing"}{"id": 5, "name": "Cross"}]
我用来生成第一个输出的代码是:
<?php
define('servername','localhost');
define('username','root');
define('password','');
define('dbname','integration');
// Create connection
$conn = new mysqli(servername, username, password, dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT id, name FROM capability_level1";
$result = $conn->query($sql);
$test = array();
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$test[] = $row["id"] . $row["name"];
}
echo json_encode($test);
} else {
echo json_encode("0 results");
}
$conn->close();
?>
我需要改变什么?这个 echo 需要在第二步传递给 ajax
最佳答案
换行
while($row = $result->fetch_assoc()) {
$test[] = $row["id"] . $row["name"];
}
到
while($row = $result->fetch_assoc()) {
$test[] = array(
'id' => $row["id"],
'name' => $row["name"]
);
}
希望这对您有所帮助。
关于php echo json_encode from DB entries -> 获取奇怪的格式。如何获得正确的 JSON,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/50854884/