如何从用PHP编写的Web服务输出JSONArray

mok*_*211 1 php json web-services

我有用PHP编写的webservice,它从本地数据库读取并以JSON格式输出结果.

但是,我无法将其输出到JSONArray中.

这是php脚本

<?php
$username = "root";
$password = "";
$hostname = "localhost"; 

$response=array();

//connection to the database
$dbhandle = mysql_connect($hostname, $username, $password) 
or die("Unable to connect to MySQL");

//select a database to work with
$selected = mysql_select_db("test",$dbhandle) 
or die("Could not select test");

//execute the SQL query and return records
$result = mysql_query("SELECT name, country FROM android");

$response["infos"] = array();   

while ($row = mysql_fetch_assoc($result)) {

    $info = array();
$info["name"]=$row["name"];
$info["country"]=$row["country"];

    print(json_encode($info));

}

//close the connection
mysql_close($dbhandle);
?>
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这是webservice的输出

{"name":"develop","country":"mru"}{"name":"fufu","country":"tutu"}  {"name":"chikaka","country":"aceVentura"}
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但我被告知这不在JSONArray中.

我在这里错过了什么?

谢谢

dig*_*rld 7

在您的示例中,您回显了多个JSON字符串,因为您的输出代码在while循环中.JSON字符串应该只有一个输出.下面的代码将为您提供JSON格式的二维数组.

$info = array();

while ($row = mysql_fetch_assoc($result)) 
{
     $arr = array();
     $arr["name"] = $row["name"];
     $arr["country"] = $row["country"];
     $info[] = $arr;
}

echo json_encode($info);
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