将数据从 MySql 编码为 JSON

Question

我的域上有 mysql 数据库在线，我有一个名为“Locations”的简单表，其中包含一些数据，我想调用数组中的数据，然后用 json 对数组进行编码，最后回显编码的 json 数据。这是我的代码，我的问题是将数据放入数组并将其编码为jso：

<?php
// obtaining connection : successful!
$con = mysql_connect("myhost", "username", "password");
$con or die(mysql_error());
mysql_select_db("database_name") or die(mysql_error());

// selecting all from table "locations" : successful!
$result = mysql_query("SELECT * FROM Locations") 
or die(mysql_error()); 

// puting data into array and encode it with json : FAILED! 
$resultArray = array();
$tempArray = array();

// Loop through each row in the result set
while($row = $result->fetch_object())
{
    // Add each row into our results array
    $tempArray = $row;
    array_push($resultArray, $tempArray);
}

// Finally, encode the array to JSON and output the results
echo json_encode($resultArray);

// closing connection : successful!
mysql_close($con);
?>

我运行我的代码，得到的结果如下：

谢谢你们。

已编辑：

我换成了mysqli，现在是这样的：

<?php

// Create connection
$con=mysqli_connect("host","user","pass","dbname");

// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

// This SQL statement selects ALL from the table 'Locations'
$sql = "SELECT * FROM Locations";

// Check if there are results
if ($result = mysqli_query($con, $sql))
{
// If so, then create a results array and a temporary one
// to hold the data
$resultArray = array();
$tempArray = array();

// Loop through each row in the result set
while($row = $result->fetch_object())
{
    // Add each row into our results array
    $tempArray = $row;
    array_push($resultArray, $tempArray);
}

// Finally, encode the array to JSON and output the results
echo json_encode($resultArray);
}

// Close connections
mysqli_close($result);
mysqli_close($con);
?>

仍然得到完全相同的错误

Answer 1

请尝试以下代码：我认为这对你有用。

$con = mysql_connect("host", "username", "password");
$con or die(mysql_error());
mysql_select_db("database_name") or die(mysql_error());

// selecting all from table "locations" : successful!
$result = mysql_query("SELECT * FROM Location") 
or die(mysql_error()); 

// puting data into array and encode it with json : FAILED! 
$resultArray = array();
$tempArray = array();

// Loop through each row in the result set
while($row = mysql_fetch_assoc($result))
{
    // Add each row into our results array
    $tempArray = $row;
    array_push($resultArray, $tempArray);
}

// Finally, encode the array to JSON and output the results
echo json_encode($resultArray);

// closing connection : successful!
mysql_close($con);

Answer 2

改变这一行：

while($row = $result->fetch_object())

到：

while($row = mysql_fetch_assoc($result))