如何：正确使用PHP将数据编码为JSON格式，并使用jquery/ajax请求数据

我正在尝试创建一个类似的地址簿

我可以成功地连接到数据库并使用php脚本插入数据

我甚至成功地显示了表行的json编码数据，尽管我不知道是否正确

我真正想要实现的目标：

$host = "localhost";
$user = "user";
$pass = "pass";
$db = "data_base";

$mysqli = new mysqli($host, $user, $pass, $db);

if($mysqli->connect_error) 
 die('Connect Error (' . mysqli_connect_errno() . ') '. mysqli_connect_error());

return $mysqli;

    id, user, pass, nickname, address, facebook, twitter, linkedin, youtube
    ID should be unique
    User is an index
    Pass is an index
    nickname is an index
    address is primary - though its possible that id should be...
    Facebook, Twitter, Linkedin, and Youtube are all indexes. 

error_reporting(E_ALL); ini_set("display_errors", 1);
include 'db/dbcon.php';
//Start connection with SQL
$q = "SELECT * FROM `cfaddrbook` WHERE key = '111111'";
$res = $mysqli->query($q) or trigger_error($mysqli->error."[$q]");
$array = array(); // initialize
while($row = $res->fetch_array(MYSQLI_BOTH)) {
$array[] = array(
'key' => $row[0],
'username' => $row[1],
// ... continue like this
);
}
header('Content-Type: application/json');
echo json_encode($array);
$res->free();
$mysqli->close();

$(document).ready(function(){ 
$.ajax({
        type: "POST",
        url: "queries.php",
        dataType: 'json',
        data: "",
        cache: false,
        success: function(result)
            {
                var cfkey = result[0];
                var user = result[1];
                alert("cfkey:" + cfkey + "user:" + user);
            }
    });
});

我希望能够对say和ID发出ajax请求，然后返回该ID对应的所有数据（用Json包装——至少我认为它需要是…）

使用ajax脚本，我希望能够将返回的相应数据保存到html文件中的输入字段中

我还想知道是否最好尝试将HTML返回到ajax调用，并以这种方式将数据输入HTML输入字段

到目前为止，我取得的成功有限，但以下是我迄今为止取得的成果

我有一个数据库连接脚本：

$host = "localhost";
$user = "user";
$pass = "pass";
$db = "data_base";

$mysqli = new mysqli($host, $user, $pass, $db);

if($mysqli->connect_error) 
 die('Connect Error (' . mysqli_connect_errno() . ') '. mysqli_connect_error());

return $mysqli;

    id, user, pass, nickname, address, facebook, twitter, linkedin, youtube
    ID should be unique
    User is an index
    Pass is an index
    nickname is an index
    address is primary - though its possible that id should be...
    Facebook, Twitter, Linkedin, and Youtube are all indexes. 

error_reporting(E_ALL); ini_set("display_errors", 1);
include 'db/dbcon.php';
//Start connection with SQL
$q = "SELECT * FROM `cfaddrbook` WHERE key = '111111'";
$res = $mysqli->query($q) or trigger_error($mysqli->error."[$q]");
$array = array(); // initialize
while($row = $res->fetch_array(MYSQLI_BOTH)) {
$array[] = array(
'key' => $row[0],
'username' => $row[1],
// ... continue like this
);
}
header('Content-Type: application/json');
echo json_encode($array);
$res->free();
$mysqli->close();

$(document).ready(function(){ 
$.ajax({
        type: "POST",
        url: "queries.php",
        dataType: 'json',
        data: "",
        cache: false,
        success: function(result)
            {
                var cfkey = result[0];
                var user = result[1];
                alert("cfkey:" + cfkey + "user:" + user);
            }
    });
});

具有以下列的mysql ISAM数据库：

$host = "localhost";
$user = "user";
$pass = "pass";
$db = "data_base";

$mysqli = new mysqli($host, $user, $pass, $db);

if($mysqli->connect_error) 
 die('Connect Error (' . mysqli_connect_errno() . ') '. mysqli_connect_error());

return $mysqli;

    id, user, pass, nickname, address, facebook, twitter, linkedin, youtube
    ID should be unique
    User is an index
    Pass is an index
    nickname is an index
    address is primary - though its possible that id should be...
    Facebook, Twitter, Linkedin, and Youtube are all indexes. 

error_reporting(E_ALL); ini_set("display_errors", 1);
include 'db/dbcon.php';
//Start connection with SQL
$q = "SELECT * FROM `cfaddrbook` WHERE key = '111111'";
$res = $mysqli->query($q) or trigger_error($mysqli->error."[$q]");
$array = array(); // initialize
while($row = $res->fetch_array(MYSQLI_BOTH)) {
$array[] = array(
'key' => $row[0],
'username' => $row[1],
// ... continue like this
);
}
header('Content-Type: application/json');
echo json_encode($array);
$res->free();
$mysqli->close();

$(document).ready(function(){ 
$.ajax({
        type: "POST",
        url: "queries.php",
        dataType: 'json',
        data: "",
        cache: false,
        success: function(result)
            {
                var cfkey = result[0];
                var user = result[1];
                alert("cfkey:" + cfkey + "user:" + user);
            }
    });
});

注：如果有人认为合适，我很乐意更改索引、主索引等

编辑**现在我的查询页面：

$host = "localhost";
$user = "user";
$pass = "pass";
$db = "data_base";

$mysqli = new mysqli($host, $user, $pass, $db);

if($mysqli->connect_error) 
 die('Connect Error (' . mysqli_connect_errno() . ') '. mysqli_connect_error());

return $mysqli;

    id, user, pass, nickname, address, facebook, twitter, linkedin, youtube
    ID should be unique
    User is an index
    Pass is an index
    nickname is an index
    address is primary - though its possible that id should be...
    Facebook, Twitter, Linkedin, and Youtube are all indexes. 

error_reporting(E_ALL); ini_set("display_errors", 1);
include 'db/dbcon.php';
//Start connection with SQL
$q = "SELECT * FROM `cfaddrbook` WHERE key = '111111'";
$res = $mysqli->query($q) or trigger_error($mysqli->error."[$q]");
$array = array(); // initialize
while($row = $res->fetch_array(MYSQLI_BOTH)) {
$array[] = array(
'key' => $row[0],
'username' => $row[1],
// ... continue like this
);
}
header('Content-Type: application/json');
echo json_encode($array);
$res->free();
$mysqli->close();

$(document).ready(function(){ 
$.ajax({
        type: "POST",
        url: "queries.php",
        dataType: 'json',
        data: "",
        cache: false,
        success: function(result)
            {
                var cfkey = result[0];
                var user = result[1];
                alert("cfkey:" + cfkey + "user:" + user);
            }
    });
});

现在，上面的脚本似乎工作正常。至少在浏览器中加载php页面时，它显示得很好

但是当我用这个脚本调用ajax时：

$host = "localhost";
$user = "user";
$pass = "pass";
$db = "data_base";

$mysqli = new mysqli($host, $user, $pass, $db);

if($mysqli->connect_error) 
 die('Connect Error (' . mysqli_connect_errno() . ') '. mysqli_connect_error());

return $mysqli;

    id, user, pass, nickname, address, facebook, twitter, linkedin, youtube
    ID should be unique
    User is an index
    Pass is an index
    nickname is an index
    address is primary - though its possible that id should be...
    Facebook, Twitter, Linkedin, and Youtube are all indexes. 

error_reporting(E_ALL); ini_set("display_errors", 1);
include 'db/dbcon.php';
//Start connection with SQL
$q = "SELECT * FROM `cfaddrbook` WHERE key = '111111'";
$res = $mysqli->query($q) or trigger_error($mysqli->error."[$q]");
$array = array(); // initialize
while($row = $res->fetch_array(MYSQLI_BOTH)) {
$array[] = array(
'key' => $row[0],
'username' => $row[1],
// ... continue like this
);
}
header('Content-Type: application/json');
echo json_encode($array);
$res->free();
$mysqli->close();

$(document).ready(function(){ 
$.ajax({
        type: "POST",
        url: "queries.php",
        dataType: 'json',
        data: "",
        cache: false,
        success: function(result)
            {
                var cfkey = result[0];
                var user = result[1];
                alert("cfkey:" + cfkey + "user:" + user);
            }
    });
});

加载此代码后，chrome控制台声明服务器返回错误500

再说一遍，我要完成的是：

$host = "localhost";
$user = "user";
$pass = "pass";
$db = "data_base";

$mysqli = new mysqli($host, $user, $pass, $db);

if($mysqli->connect_error) 
 die('Connect Error (' . mysqli_connect_errno() . ') '. mysqli_connect_error());

return $mysqli;

    id, user, pass, nickname, address, facebook, twitter, linkedin, youtube
    ID should be unique
    User is an index
    Pass is an index
    nickname is an index
    address is primary - though its possible that id should be...
    Facebook, Twitter, Linkedin, and Youtube are all indexes. 

error_reporting(E_ALL); ini_set("display_errors", 1);
include 'db/dbcon.php';
//Start connection with SQL
$q = "SELECT * FROM `cfaddrbook` WHERE key = '111111'";
$res = $mysqli->query($q) or trigger_error($mysqli->error."[$q]");
$array = array(); // initialize
while($row = $res->fetch_array(MYSQLI_BOTH)) {
$array[] = array(
'key' => $row[0],
'username' => $row[1],
// ... continue like this
);
}
header('Content-Type: application/json');
echo json_encode($array);
$res->free();
$mysqli->close();

$(document).ready(function(){ 
$.ajax({
        type: "POST",
        url: "queries.php",
        dataType: 'json',
        data: "",
        cache: false,
        success: function(result)
            {
                var cfkey = result[0];
                var user = result[1];
                alert("cfkey:" + cfkey + "user:" + user);
            }
    });
});

我希望能够对say和ID发出ajax请求，然后返回该ID对应的所有数据（用Json包装——至少我认为它需要是…）

使用ajax脚本，我希望能够将返回的相应数据保存到html的输入字段中

编辑： 最后我发现我和Majid讨论的问题是SQL查询。

---

密钥需要用`字符'包装。

我相信您希望querys.php返回json（到ajax），因此您需要querys.php中的内容头类型

header('Content-Type: application/json');

---

您需要更多有用的错误消息。尝试在代码开头添加以下行

错误报告（E_全部）；

---

ini设置（“显示错误”，1）

输出JSON的脚本正在编写多个有效的JSON字符串（每个数据库行一个），但这些字符串加起来并不是一个有效的JSON文件。JSON文件应表示一个JSON对象

如果要传递一个ID并返回一个数据库行，则必须将该ID添加到AJAX调用的

数据部分，并修改querys.php以将该ID从其
$\u POST
数组传递到MySQL查询的
WHERE
部分。然后，只输出一个JSON编码的对象，而不是多个，这将是一个有效的JSON文件

（或者，如果希望返回整个表，可以
json_encode（）
整个
$rows
数组，而不是每个
$row
）

另外，如果您使用PHP中的字符串索引数组
json\u encode（）
，则可以通过名称而不是索引在Javascript中读取其属性。您已经经历了用PHP命名键的麻烦，然后切换回使用Javascript中基于0的索引来引用它们。你可以选择一种或另一种方式，但你只能选择一种
 执行查询后，结果集在
$res
中可用，您只需构建数组，无需单独的foreach：

$array = array(); // initialize
while($row = $res->fetch_array(MYSQLI_BOTH)) {
  $array[] = array(
    'id'       => $row[0],
    'username' => $row[1],
    'password' => $row[2],
    'nick'     => $row[3],
    'addr'     => $row[4],
    'facebook' => $row[5],
    'twitter'  => $row[6],
    'linkedin' => $row[7],
    'youtube'  => $row[8]
  );
}
header('Content-Type: application/json');
echo json_encode($array);

还请注意，这样，您的json将具有键，因此要使用它，您应该更改：

success: function(result) {
  var cfkey = result[0];
  var user = result[1];
  alert("cfkey:" + cfkey + "user:" + user);
}

到

或者干脆

$.getJSON('queries.php', {cfkey: $("#cfkey").val()}, function(result) {
  // we have multiple results
  $.each(result, function(i,r) {
    console.log("cfkey:" + r.key + "user:" + r.username);
  });
});

编辑：添加了@amurrell
指出的标题，在连接脚本中，
$pass
之后，您缺少了一个
“
。当我拿出真正的密码将代码粘贴到这里时，这是一个输入错误。我服务器上的实际代码实际上在php代码中正确地说明了$pass=”密码。但是，谢谢您！我在下面添加了它。”include”行，控制台再次返回了500错误（对于ajax查询）。我是否将其添加到了错误的位置？您可能应该将其放在include之前。在PHP的必须在发送任何实际输出之前调用，可以通过普通HTML标记、文件中的空行，也可以从PHP调用。使用include或require函数或其他文件访问函数读取代码，并且在header（）之前输出空格或空行，这是非常常见的错误调用。使用单个PHP/HTML文件时也存在相同的问题。它现在位于包含行上方。仍然没有区别。您的回答与我试图实现的基本相同。请您详细说明第1、2和3段，好吗？因此，如果您将
数据
设置为
{id:1}
，那么在PHP中，

$\u POST['id']将是

1

。现在，只需将MySQL查询硬编码为

SELECT*FROM cfaddrbook，其中id=1

（尽管选择一个现有的id号），您将看到返回有效的JSON。在成功处理程序中，更改

结果[0]

到

结果.id

和

结果[1]

到

结果。username

，它应该按照您想要的方式工作。您应该能够通过更多的实验从中找到答案。祝您好运！在添加WHERE id=1行之后，在我的例子WHERE key=111111中，它会吐出很多错误。如果我这样做，在显示php页面时不会再有错误。另外，addURL后的ng&username仍然显示所有JSON数据，而不是对应的特定数据。因此我想知道，当数据被传递回ajax查询时，它是否找不到其请求？非常棒的建议！但是，输入代码后，php给了我：太多的编辑！抱歉。这表明我该睡觉了！”PHP分析错误：语法错误，意外“，”在/file/。。。在线8@user3259138是的，我有