ajax 到 php 并从中获取 JSON
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ajax to a php and get JSON from it
提问by pencilvania
I have a php file that i connect to it with ajax and callback value is JSON when i get data from php it dosnt show and when alert data i see Object
我有一个 php 文件，我用 ajax 连接到它，当我从 php 获取数据时它不显示，当我看到对象时，回调值是 JSON
Where is my problem ?
我的问题在哪里？
PHP:
PHP:
if(isset($_SERVER["HTTP_X_REQUESTED_WITH"])){
$query = mysql_query("select * from tab");
for ($i=0;$i<mysql_num_rows($query);$i++){
while($row = mysql_fetch_assoc($query)){
$title['t'][i] = $row['title'];
$title['d'][i] = $row['description'];
}
}
echo(json_encode($title));
exit();
?>
JS:
JS：
$('#button').click(function(){
$.ajax({
url : "test2.php",
data : $("#tab"),
type : "GET",
success : function(b){
b = eval('('+ b +')');
console.log((b['t']));
alert(b);
}
});
});
How can i get all of data from this JSON and show me corect it ?
我怎样才能从这个 JSON 中获取所有数据并向我展示它的正确性？
回答by Jonast92
Here's a full working example with single row fetch and multi row fetch, without using mysql_ syntax and using prepared statements to prevent sql injections.
这是一个完整的工作示例，包含单行提取和多行提取，不使用 mysql_ 语法并使用准备好的语句来防止 sql 注入。
And yes, DON'T use mysql specific syntax, like I mentioned here: I cant get the form data to go into database. What am I doing wrong?
是的，不要使用特定于 mysql 的语法，就像我在这里提到的那样：我无法将表单数据导入数据库。我究竟做错了什么？
function example()
{
var select = true;
var url = '../scripts/ajax.php';
$.ajax(
{
// Post select to url.
type : 'post',
url : url,
dataType : 'json', // expected returned data format.
data :
{
'select' : select // the variable you're posting.
},
success : function(data)
{
// This happens AFTER the PHP has returned an JSON array,
// as explained below.
var result1, result2, message;
for(var i = 0; i < data.length; i++)
{
// Parse through the JSON array which was returned.
// A proper error handling should be added here (check if
// everything went successful or not)
result1 = data[i].result1;
result2 = data[i].result2;
message = data[i].message;
// Do something with result and result2, or message.
// For example:
$('#content').html(result1);
// Or just alert / log the data.
alert(result1);
}
},
complete : function(data)
{
// do something, not critical.
}
});
}
Now we need to receive the posted variable in ajax.php:
现在我们需要在ajax.php中接收posted变量：
$select = isset($_POST['select']) ? $_POST['select'] : false;
The ternary operator lets $select's value become false if It's not set.
如果未设置，三元运算符会让 $select 的值变为 false。
Make sure you got access to your database here:
确保您可以在此处访问您的数据库：
$db = $GLOBALS['db']; // An example of a PDO database connection
Now, check if $select is requested (true) and then perform some database requests, and return them with JSON:
现在，检查是否请求了 $select (true)，然后执行一些数据库请求，并使用 JSON 返回它们：
if($select)
{
// Fetch data from the database.
// Return the data with a JSON array (see below).
}
else
{
$json[] = array
(
'message' => 'Not Requested'
);
}
echo json_encode($json);
flush();
How you fetch the data from the database is of course optional, you can use JSON to fetch a single row from the database or you can use it return multiple rows.
从数据库中获取数据的方式当然是可选的，您可以使用 JSON 从数据库中获取单行，也可以使用它返回多行。
Let me give you an example of how you can return multiple rows with json (which you will iterate through in the javascript (the data)):
让我举一个例子，说明如何使用 json 返回多行（您将在 javascript（数据）中对其进行迭代）：
function selectMultipleRows($db, $query)
{
$array = array();
$stmt = $db->prepare($query);
$stmt->execute();
if($result = $stmt->fetchAll(PDO::FETCH_ASSOC))
{
foreach($result as $res)
{
foreach($res as $key=>$val)
{
$temp[$key] = utf8_encode($val);
}
array_push($array, $temp);
}
return $array;
}
return false;
}
Then you can do something like this:
然后你可以做这样的事情：
if($select)
{
$array = array();
$i = 0;
$query = 'SELECT e.result1, e.result2 FROM exampleTable e ORDER BY e.id ASC;';
foreach(selectMultipleRows($db, $query) as $row)
{
$array[$i]["result1"] = $row['result1'];
$array[$i]["result2"] = $row['result2'];
$i++;
}
if(!(empty($array))) // If something was fetched
{
while(list($key, $value) = each($array))
{
$json[] = array
(
'result1' => $value["result1"],
'result2' => $value["result2"],
'message' => 'success'
);
}
}
else // Nothing found in database
{
$json[] = array
(
'message' => 'nothing found'
);
}
}
// ...
Or, if you want to KISS (Keep it simple stupid):
或者，如果你想亲吻（保持简单愚蠢）：
Init a basic function which select some values from the database and returns a single row:
初始化一个基本函数，它从数据库中选择一些值并返回单行：
function getSingleRow($db, $query)
{
$stmt = $db->prepare($query);
$stmt->execute();
// $stmt->execute(array(":id"=>$someValue)); another approach to execute.
$result = $stmt->fetch(PDO::FETCH_ASSOC);
if($result)
{
$array = (
'result1' => $result['result1'],
'result2' => $result['result2']
);
// An array is not needed for a single value.
return $array;
}
return false;
}
And then fetch the row (or the single value) and return it with JSON:
然后获取行（或单个值）并用 JSON 返回它：
if($select)
{
// Assume that the previously defined query exists.
$results = getSingleRow($db, $query);
if($results !== false)
{
$json[] = array
(
'result1' => $results['result1'],
'result2' => $results['result2'],
'message' => 'success'
);
}
else // Nothing found in database
{
$json[] = array
(
'message' => 'nothing found'
);
}
}
// ...
And if you want to get the value of $("#tab") then you have to do something like $("#tab").val() or $("#tab").text().
如果你想获得 $("#tab") 的值，那么你必须做一些类似 $("#tab").val() 或 $("#tab").text() 的事情。
I hope that helps.
我希望这有帮助。
