PHP传给前端的值有大量html代码

话不多说, 直接上代码

<br />
<font size="1"><table class="xdebug-error xe-deprecated" dir="ltr" border="1" cellspacing="0" cellpadding="1">
<tr><th align="left" bgcolor="#f57900" colspan="5"><span style="background-color: #cc0000; color: #fce94f; font-size: x-large;">( ! )</span> Deprecated: mysql_connect(): The mysql extension is deprecated and will be removed in the future: use mysqli or PDO instead in C:wampwwwunit1json1.php on line <i>5</i></th></tr>
<tr><th align="left" bgcolor="#e9b96e" colspan="5">Call Stack</th></tr>
<tr><th align="center" bgcolor="#eeeeec">#</th><th align="left" bgcolor="#eeeeec">Time</th><th align="left" bgcolor="#eeeeec">Memory</th><th align="left" bgcolor="#eeeeec">Function</th><th align="left" bgcolor="#eeeeec">Location</th></tr>
<tr><td bgcolor="#eeeeec" align="center">1</td><td bgcolor="#eeeeec" align="center">0.0000</td><td bgcolor="#eeeeec" align="right">242216</td><td bgcolor="#eeeeec">{main}(  )</td><td title="C:wampwwwunit1json1.php" bgcolor="#eeeeec">..json1.php<b>:</b>0</td></tr>
<tr><td bgcolor="#eeeeec" align="center">2</td><td bgcolor="#eeeeec" align="center">0.0000</td><td bgcolor="#eeeeec" align="right">242688</td><td bgcolor="#eeeeec"><a href="http://www.php.net/function.mysql-connect" target="_new">mysql_connect</a>
(  )</td><td title="C:wampwwwunit1json1.php" bgcolor="#eeeeec">..json1.php<b>:</b>5</td></tr>
</table></font>
[{"0":"Test1","product":"Test1","1":"Hello","questions":"Hello"},{"0":"","product":"","1":"","questions":""},{"0":"Test1","product":"Test1","1":"Venky","questions":"Venky"},{"0":"","product":"","1":"","questions":""}]

其实我们本意只想传下面的json数据

[{"0":"Test1","product":"Test1","1":"Hello","questions":"Hello"},{"0":"","product":"","1":"","questions":""},{"0":"Test1","product":"Test1","1":"Venky","questions":"Venky"},{"0":"","product":"","1":"","questions":""}]

那上面多出来的html代码都是什么鬼?
那就再来看下PHP代码

<?php

header("Content-type:application/json");

mysql_connect("localhost","root","") or die(mysql_error()); 

mysql_select_db("testdb");

$select = mysql_query("select * from questions");

$rows = array();

while($row=mysql_fetch_array($select))
{

    $rows[] = $row;
}

echo json_encode($rows);

?>

返回的json字符串中, 有个关键词, “Deprecated”, 代表 “不建议使用”.PHP给出的解释是: mysql_connect() 这个方法在将来的某个时间会被彻底移除掉.
而且, 在声明一点, 返回的json数据走的是error方法,至于为什么会返回正确的数据,这就得问大神了.

解决方案如下
使用PDO或者mysqli_connect();
PDO代码:

<?php
//  header("Content-type:application/json");
    // 连接数据库
    $dsn = "mysql:dbname=dbtest1;host=127.0.0.1";
    $user = "root";
    $password = "";

    try {
$dbh = new PDO("mysql:dbname=dbtest1;host=127.0.0.1", $user, $pass);
$dbh = new PDO($dsn, $user, $password);
        foreach($dbh->query("SELECT * from tb_test1") as $row) {
            print_r($row[0]);
            print_r("
");
        }
        $dbh = null;
    } catch (PDOException $e) {
        print "Error!: " . $e->getMessage() . "<br/>";
        die();
    }
?>

我自己用的是PDO,本以为能好了, 结果还是报类似的错: 一大堆html代码+返回数据
后来把PDO对象的第一个参数由常量改为变量后, 就好了, 如下:

$dbh = new PDO($dsn, $user, $pass);

公司同事给出了一个猜测性解释——PDO实例在后续操作中可能会影响到参数改变, 常量不可变, 变量可变, 所以常量填进去后会报错.
谨以此文,抛砖引玉,希望大神来探讨.