Php 按JSON数组对SQL数据进行分组
我正在为教师制作一个页面，显示学生在考试中的成绩 要显示图形，我想使用Php 按JSON数组对SQL数据进行分组,php,mysql,json,pdo,highcharts,Php,Mysql,Json,Pdo,Highcharts,我正在为教师制作一个页面，显示学生在考试中的成绩 要显示图形，我想使用HighchartsJavaScript库 到目前为止，我有一个PHP脚本，它使用PDO创建JSON数据，以后可以从不同的页面将这些数据馈送到Highcharts 我的问题： 如何将来自同一学生的所有数据分组到一个数组中？关于我希望实现的目标，请参见最后一个示例。另外：我希望将所有数据封装在一个总体JSON数组中 我想要这个： [{ "student": "Andreas", "level" : [4, 3] }, {
HighchartsPHPPDOHighcharts学生的所有数据分组到一个数组中？关于我希望实现的目标，请参见最后一个示例。另外：我希望将所有数据封装在一个总体JSON数组中
我想要这个：
[{
"student": "Andreas",
"level" : [4, 3]
}, {
"student": "Eivind",
"level" : [4, 5]
}, {
"student": "Ole",
"level" : [4, 3]
}]
<?php
require("config.inc.php");
$school = $_GET["school"];
$class = $_GET["class"];
//initial query
$query = 'SELECT student, taskid, level FROM task
WHERE school=' . '"' . $school . '"' . ' AND class=' . '"' . $class . '" ORDER BY student';
//execute query
try {
$stmt = $db->prepare($query);
$result = $stmt->execute();
}
catch (PDOException $ex) {
$response["success"] = 0;
$response["message"] = "Database Error!";
die(json_encode($response));
}
// Finally, we can retrieve all of the found rows into an array using fetchAll
$rows = $stmt->fetchAll();
if ($rows) {
$response["posts"] = array();
foreach ($rows as $row) {
$post = array();
$post["student"] = $row["student"];
$post["level"] = $row["level"];
//update our repsonse JSON data
array_push($response["posts"], $post);
}
// echoing JSON response
echo json_encode($response, JSON_NUMERIC_CHECK);
} else {
$response["success"] = 0;
$response["message"] = "No Post Available!";
die(json_encode($response));
}
?>
{
"posts": [
{
"student": "Andreas",
"level": 4
},
{
"student": "Andreas",
"level": 3
},
{
"student": "Eivind",
"level": 4
},
{
"student": "Eivind",
"level": 5
},
{
"student": "Ole",
"level": 4
},
{
"student": "Ole",
"level": 3
}
]
}
这就是我的PHP的样子：
[{
"student": "Andreas",
"level" : [4, 3]
}, {
"student": "Eivind",
"level" : [4, 5]
}, {
"student": "Ole",
"level" : [4, 3]
}]
<?php
require("config.inc.php");
$school = $_GET["school"];
$class = $_GET["class"];
//initial query
$query = 'SELECT student, taskid, level FROM task
WHERE school=' . '"' . $school . '"' . ' AND class=' . '"' . $class . '" ORDER BY student';
//execute query
try {
$stmt = $db->prepare($query);
$result = $stmt->execute();
}
catch (PDOException $ex) {
$response["success"] = 0;
$response["message"] = "Database Error!";
die(json_encode($response));
}
// Finally, we can retrieve all of the found rows into an array using fetchAll
$rows = $stmt->fetchAll();
if ($rows) {
$response["posts"] = array();
foreach ($rows as $row) {
$post = array();
$post["student"] = $row["student"];
$post["level"] = $row["level"];
//update our repsonse JSON data
array_push($response["posts"], $post);
}
// echoing JSON response
echo json_encode($response, JSON_NUMERIC_CHECK);
} else {
$response["success"] = 0;
$response["message"] = "No Post Available!";
die(json_encode($response));
}
?>
{
"posts": [
{
"student": "Andreas",
"level": 4
},
{
"student": "Andreas",
"level": 3
},
{
"student": "Eivind",
"level": 4
},
{
"student": "Eivind",
"level": 5
},
{
"student": "Ole",
"level": 4
},
{
"student": "Ole",
"level": 3
}
]
}
您将获得以下输出：
[{"category":"Search Engines","hits":13,"bytes":673684},
{"category":"Content Server","hits":3,"bytes":88930},
{"category":"Internet Services","hits":1,"bytes":3690},
{"category":"Business","hits":1,"bytes":2847}]
在php中，您可以执行以下操作：
在你的面前
foreach ($rows as $row) {
$post = array();
$post["student"] = $row["student"];
$post["level"] = $row["level"];
//update our repsonse JSON data
array_push($response["posts"], $post);
}
您需要这样做：
<?php
$response = array();
$response["posts"] = array();
$students = array("student1", "student2", "student3", "student1");
$tempArray = array();
foreach ($students as $student) {
$lvl = 1;
if(!isset($tempArray[$student])){
$tempArray[$student] = array("name" => $student, "level" => array($lvl));
}
else{
$tempArray[$student]["level"][] = $lvl;
}
}
// add it to the array
$response["posts"] = $tempArray;
// encode array
$response = json_encode($response);
echo "<br><br><br> ";
// example
// decode and make it a array for easier looping.
$response = (array)json_decode($response);
$responsePosts = (array) $response["posts"];
// foreach post
foreach($response["posts"] as $keyP => $valueP){
// convert to array same as above
$valueP = (array) $valueP;
// key that is kinda useless but made it the student name so i can easily see if it already exists
echo "key :".$keyP." <br />";
// name of the student
echo "name :".$valueP["name"]." <br />";
// all the levels
echo "levels: ";
foreach($valueP["level"] as $lvl){
echo $lvl." ";
}
echo "<br /><br />";
}
?>
这应该行得通。Atm我使用学生名作为一个键，以查看它是否已经存在，否则您需要遍历所有数组并执行“name”=$student最有可能的操作
不需要推送。通过使用学生姓名作为数组键，您可以非常轻松地构建这样的数组。构建数组后，可以使用将字符串键转换回数字键
...
if ($rows) {
foreach ($rows as $row) {
$posts[$row['student']]['student'] = $row['student'];
$posts[$row['student']]['level'][] = $row['level'];
}
$response = array_values($posts);
echo json_encode($response, JSON_NUMERIC_CHECK);
} else { ...
这将为您提供以下$response
：
[
{
"student": "Andreas",
"level": [4, 3]
},
{
"student": "Eivind",
"level": [4, 5]
},
{
"student": "Ole",
"level": [4, 3]
}
]
谢谢我添加了array\u push（$response[“posts”]，$tempArray）
在foreach的末尾，我得到的结果是：接近，但还没有到达。。如果我不在foreach
的末尾执行array\u push
，JSON数据返回器将为空。我将看看是否可以在我的示例中包含push，并使其与您要求的完全相同。我推荐@Prashant给我的答案，如果你能做到的话。这个应该可以。要获得所需的精确数组，可以执行var_dump（$response[“posts”][0]）；数组推送创建了一个键0，我现在不知道如何删除它。我有点傻。我删除了数组并将其分配给了数组。这正是你要找的。你在最近几次编辑中把我弄丢了。。需要对´´$students´´进行硬编码，因此，如果这些数据是由实际学生生成的，则没有任何用处：-（另外，在你最新的方法中，我没有得到JSON编码的数据，只是得到了一些PHP转储。谢谢你的努力，但到目前为止编辑#2似乎最接近。这太棒了！谢谢你。