下面的代码在这里可以正常工作!

<?php

  $con=mysqli_connect("localhost",$username,$password,databaseName);

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

  $query = "the query here";

  $result = mysqli_query($con,$query);

  $rows = array();
  while($r = mysqli_fetch_array($result)) {
    $rows[] = $r;
  }
  echo json_encode($rows);

  mysqli_close($con);
?>

使用FOR循环的另一个选项:

 $sth = mysql_query("SELECT ...");
 for($rows = array(); $row = mysql_fetch_assoc($sth); $rows[] = $row);
 print json_encode($rows);

唯一的缺点是循环for比while或特别是foreach慢

考虑到在mysql中通常没有任何嵌套的json对象，创建自己的编码函数是相当容易的

首先，检索mysqli的函数会生成一个数组:

function noom($rz) {
    $ar = array();
    if(mysqli_num_rows($rz) > 0) {
        while($k = mysqli_fetch_assoc($rz)) {
            foreach($k as $ki=>$v) {
                $ar[$ki] = $v;
            }
        }
    }
    return $ar;
}

现在，函数将数组编码为json:

function json($ar) {
    $str = "";
    $str .= "{";
    $id = 0;
    foreach($ar as $a=>$b) {
        $id++;
        $str .= "\"".$a."\":";
        if(!is_numeric($b)) {
            $str .= "\"".$b."\"";
        } else {
            $str .= $b;
        }
        
        if($id < count($ar)) {
            $str .= ",";
        }
    }
    $str .= "}";
    return $str;
}

然后使用它:

<?php
$o = new mysqli(
    "localhost",
    "root",""
);
if($o->connect_error) {
    echo "DUDE what are you/!";
} else {
    $rz = mysqli_query($o,
        "SELECT * FROM mydatabase.mytable"
    );
    $ar = noom($rz);
    echo json($ar);
}
?>

例如 $result = mysql_query("SELECT * FROM userprofiles where NAME='TESTUSER' ");

1.)如果$result只有一行。

$response = mysql_fetch_array($result);
echo json_encode($response);

2.)如果$result多于一行。你需要迭代这些行，并将其保存到一个数组中，并返回一个包含数组的json。

$rows = array();
if (mysql_num_rows($result) > 0) {
    while($r = mysql_fetch_assoc($result)) {
       $id = $r["USERID"];   //a column name (ex.ID) used to get a value of the single row at at time
       $rows[$id] = $r; //save the fetched row and add it to the array.
    }
}    
echo json_encode($rows);

http://www.php.net/mysql_query说“mysql_query()返回一个资源”。

http://www.php.net/json_encode说它可以编码任何值，“除了资源”。

您需要在数组中遍历和收集数据库结果，然后对数组进行json_encode。

抱歉，这是在问题之后很长一段时间，但是:

$sql = 'SELECT CONCAT("[", GROUP_CONCAT(CONCAT("{username:'",username,"'"), CONCAT(",email:'",email),"'}")), "]") 
AS json 
FROM users;'
$msl = mysql_query($sql)
print($msl["json"]);

基本上:

"SELECT" Select the rows    
"CONCAT" Returns the string that results from concatenating (joining) all the arguments
"GROUP_CONCAT" Returns a string with concatenated non-NULL value from a group