这是我的代码,但它只适用于mySQLResultsetToJSON()
中的for循环。
<?php
$servername = "127.0.0.1";
$username = "root";
$password = "123456";
$table = "ticketing";
$link = new mysqli($servername, $username, $password, $table);
if ($link->connect_error) {
die("Connection failed: " . $link->connect_error);
}
echo "Connected successfully";
$result = $link->query("SELECT * from Ticket");
print_r($result);
$final_result = mySQLResultsetToJSON($result);
print_r($final_result);
$link->close();
function mySQLResultsetToJSON($resultSet)
{
for($i = 0; sizeof($resultSet); $i++)
{
$rows = array();
while($r = mysqli_fetch_assoc($resultSet[$i])) {
$rows[] = $r;
}
$jsonResult[$i] = json_encode(array('Results' => $rows));
}
print_r($jsonResult);
return $jsonResult;
}
?>
谢谢!
托马斯
答案 0 :(得分:1)
echo "mysql data<br />";
$result = $link->query("SELECT * from users");
print_r($result->fetch_object());
echo "<br />";
echo "in json<br />";
$res = ['Results' => $result->fetch_object() ];
echo json_encode($res);
$link->close();
答案 1 :(得分:0)
用户喜欢
$result = $link->query("SELECT * from Ticket");
$rows = array();
while($r = mysqli_fetch_assoc($result)) {
$rows[] = $r;
}
print "<pre>";
print_r(json_encode(array('Results' =>$rows)));
$link->close();