如何将mysql_fetch_row解析为JSON对象

Question

我有以下代码:

<?php 
include 'db.php';
$con = mysql_connect($host,$user,$pass);
$dbs = mysql_select_db($databaseName, $con);

$tableName = "tbl_title";
$result = mysql_query("SELECT id, fld_title FROM $tableName");

$data = array();
while ( $row = mysql_fetch_row($result) )
{
    //$data[] = $row;
    //$data['item'][] = $row;
    //$data['item'][] = $row['id'].value;
    $id = $row['id'];
    $fld_title = $row['fld_title'];
    $data = array( "id" => $id, "fld_title" => $fld_title);

    //echo $row['id'] . " " . $row['fld_title'];
            //echo '<option value="'.$row->id.'">'.$row->fld_title.'</option>'; 
}
echo json_encode( $data );  
    //  echo $row[0] . " " . $row[1];
?>

我试图以格式化的方式返回结果,但是,即使我从查询中获得结果,$ id和$ fld_title也不会设置为值.我错过了什么？

Answer 1

没有必要引用每一列.$row已经是一个数组了.我已经改为使用mysql_fetch_assoc,因为那个给了命名列.

$data = array();
while ( $row = mysql_fetch_assoc($result) )
{
    $data[] = $row;
}
echo json_encode( $data );