php JSON 编码 MySQL 结果
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/383631/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
JSON encode MySQL results
提问by Trick Jarrett
How do I use the json_encode()function with MySQL query results? Do I need to iterate through the rows or can I just apply it to the entire results object?
如何将json_encode()函数与 MySQL 查询结果一起使用？我需要遍历行还是可以将其应用于整个结果对象？
回答by Paolo Bergantino
$sth = mysqli_query("SELECT ...");
$rows = array();
while($r = mysqli_fetch_assoc($sth)) {
$rows[] = $r;
}
print json_encode($rows);
The function json_encodeneeds PHP >= 5.2 and the php-jsonpackage - as mentioned here
该函数json_encode需要 PHP >= 5.2 和php-json包 - 如这里所述
NOTE: mysqlis deprecated as of PHP 5.5.0, use mysqliextension instead http://php.net/manual/en/migration55.deprecated.php.
注意：mysql自 PHP 5.5.0 起已弃用，请使用mysqli扩展名代替http://php.net/manual/en/migration55.deprecated.php。
回答by ddavtian
Try this, this will create your object properly
试试这个，这将正确创建您的对象
$result = mysql_query("SELECT ...");
$rows = array();
while($r = mysql_fetch_assoc($result)) {
$rows['object_name'][] = $r;
}
print json_encode($rows);
回答by Hugh Bothwell
http://www.php.net/mysql_querysays "mysql_query()returns a resource".
http://www.php.net/mysql_query说“mysql_query()返回资源”。
http://www.php.net/json_encodesays it can encode any value "except a resource".
http://www.php.net/json_encode说它可以编码“除了资源”的任何值。
You need to iterate through and collect the database results in an array, then json_encodethe array.
您需要遍历并将数据库结果收集到一个数组中，然后是json_encode该数组。
回答by Tokes
Thanks this helped me a lot. My code:
谢谢这对我帮助很大。我的代码：
$sqldata = mysql_query("SELECT * FROM `$table`");
$rows = array();
while($r = mysql_fetch_assoc($sqldata)) {
$rows[] = $r;
}
echo json_encode($rows);
回答by aashima
Thanks.. my answer goes:
谢谢..我的回答是：
if ($result->num_rows > 0) {
# code...
$arr = [];
$inc = 0;
while ($row = $result->fetch_assoc()) {
# code...
$jsonArrayObject = (array('lat' => $row["lat"], 'lon' => $row["lon"], 'addr' => $row["address"]));
$arr[$inc] = $jsonArrayObject;
$inc++;
}
$json_array = json_encode($arr);
echo $json_array;
}
else{
echo "0 results";
}
回答by P?r
The above will not work, in my experience, before you name the root-element in the array to something, I have not been able to access anything in the final json before that.
以上是行不通的，根据我的经验，在将数组中的根元素命名为某些东西之前，在此之前我无法访问最终 json 中的任何内容。
$sth = mysql_query("SELECT ...");
$rows = array();
while($r = mysql_fetch_assoc($sth)) {
$rows['root_name'] = $r;
}
print json_encode($rows);
That should do the trick!
这应该够了吧！
P?r
P?r
回答by ferreirabraga
The code below works fine here!
下面的代码在这里工作正常！
<?php
$con=mysqli_connect("localhost",$username,$password,databaseName);
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$query = "the query here";
$result = mysqli_query($con,$query);
$rows = array();
while($r = mysqli_fetch_array($result)) {
$rows[] = $r;
}
echo json_encode($rows);
mysqli_close($con);
?>
回答by James
My simple fix to stop it putting speech marks around numeric values...
我的简单修复是阻止它在数值周围放置语音标记......
while($r = mysql_fetch_assoc($rs)){
while($elm=each($r))
{
if(is_numeric($r[$elm["key"]])){
$r[$elm["key"]]=intval($r[$elm["key"]]);
}
}
$rows[] = $r;
}
回答by gear4
Sorry, this is extremely long after the question, but:
对不起，这是在问题之后很长一段时间，但是：
$sql = 'SELECT CONCAT("[", GROUP_CONCAT(CONCAT("{username:'",username,"'"), CONCAT(",email:'",email),"'}")), "]")
AS json
FROM users;'
$msl = mysql_query($sql)
print($msl["json"]);
Just basically:
基本上就是：
"SELECT" Select the rows
"CONCAT" Returns the string that results from concatenating (joining) all the arguments
"GROUP_CONCAT" Returns a string with concatenated non-NULL value from a group
回答by jrran90
we could simplify Paolo Bergantinoanswer like this
我们可以像这样简化Paolo Bergantino 的回答
$sth = mysql_query("SELECT ...");
print json_encode(mysql_fetch_assoc($sth));
