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php简单实现查询数据库返回json数据

PHPz
Release: 2018-10-09 17:18:45
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这篇文章主要介绍了php简单实现查询数据库返回json数据,并附上2则示例代码,非常的简单实用,有需要的小伙伴可以参考下。

示例代码一:

// 设置返回json格式数据
header('content-type:application/json;charset=utf8');
//连接数据库
$link = mysql_connect("localhost", "root", "root") or die("Unable to connect to the MySQL!");
mysql_query("SET NAMES 'UTF8'");
mysql_select_db("jilinwula", $link) or die("Unable to connect to the MySQL!");
// 获取分页参数
$page = 0 ;
$pageSize = 3;
if(!is_null($_GET["page"])) {
$page = $_GET["page"];
}
if(!is_null($_GET["pageSize"])) {
$pageSize = $_GET["pageSize"];
}
// 查询数据到数组中
$result = mysql_query("select username,password from userinfo limit " . $page . ", ". $pageSize ."");
$results = array();
while ($row = mysql_fetch_assoc($result)) {
$results[] = $row;
}
// 将数组转成json格式
echo json_encode($results);
// 关闭连接
mysql_free_result($result);
mysql_close($link);
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示例代码二:

<?php
//需要执行的SQL语句
//单条
$sql="select id,name from tbl_user where id=1";
//多条数据
//$sql="select id,name from tbl_user";
//调用conn.php文件进行数据库操作 
require(&#39;Conn.php&#39;); 
//提示操作成功信息,注意:$result存在于conn.php文件中,被调用出来 
if($result) 
{ 
// $array=mysql_fetch_array($result,MYSQL_ASSOC);
 /*数据集
 $users=array();
 $i=0;
 while($row=mysql_fetch_array($result,MYSQL_ASSOC)){
  echo $row[&#39;id&#39;].&#39;-----------&#39;.$row[&#39;name&#39;].&#39;</br>&#39;;
  $users[$i]=$row;
  $i++;
 }
 echo json_encode(array(&#39;dataList&#39;=>$users));
 */
 /*单条数据*/
 $row=mysql_fetch_row($result,MYSQL_ASSOC);
 echo json_encode(array(&#39;jsonObj&#39;=>$row));
} 
mysql_free_result($result);
//释放结果
mysql_close();
//关闭连接
?>
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 以上所述就是本文的全部内容了,希望大家能够喜欢。

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source:jb51.net
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