Question

我是php的新手。我正在努力从数据库中获取结果，并将这些以Json格式发送的详细信息发送到ajax调用。而我无法使用json_encode进行转换。我希望以

的格式获得结果

[{"id":1,"name":"Rafael","password":"rafael"},{"id":1,"name":"nadal","password":"nadal"}]

我的PHP代码是

// credentials of MySql database.
$username = "root";
$password = "admin";
$hostname = "localhost"; 

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

$selected = mysql_select_db("Angular",$dbhandle)
or die("Could not select Angular");
//execute the SQL query and return records
$result = mysql_query("SELECT name,password FROM User");

//fetch tha data from the database  
while ($row = mysql_fetch_array($result)) {

    $id = $row{'id'};
    $name = $row{'name'};
    $password = $row{'password'};
    $jsonData = json_encode($name,$password);
}
echo "Json:".$jsonData;

Answer 1

json_encode中的参数必须是数组。

你应该使用

$jsonData = json_encode( $array ); //$array is your array element

Answer 2

将mysql_query后的代码更改为：

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

  $jsonArray[] = $row;
}

echo json_encode($jsonArray);

请注意，您不要在查询中选择ID。它应该是

$result = mysql_query("SELECT id, name, password FROM User");

Answer 3

使用像这样的东西

$ queryString =＆＃34; SELECT * FROM user＆＃34 ;;

$ query = mysql_query（$ queryString）或die（mysql_error（））;

$db = array();
while($dbs = mysql_fetch_assoc($query)) {
    $db[] = $dbs;
}

echo 

$output = json_encode(array(
    "success" => mysql_errno() == 0,

    "information" => $db
));

myoutput看起来像这样;

它将显示数据库表中的所有字段

{＆＃34;成功＆＃34;：真，＆＃34;信息＆＃34;：[{＆＃34; ID＆＃34;：＆＃34; 1＆＃34;＆＃34;用户名＆＃ 34;：＆＃34; sampleName＆＃34;＆＃34;名字＆＃34;：＆＃34; SampleLastName＆＃34;}] }

希望它有效：D

将php数据转换为Json

3 个答案: