[英]From SQL table to JSON through PHP
我幾天來一直在努力解決這個問題而且找不到合適的解決方案。
我想通過PHP從這個數據庫表中檢索數據:
而且,再次使用php,有一個類似於此的JSON輸出:
[{
"label": "2013-01-07",
"value": "4"
}, {
"label": "2013-01-06",
"value": "65"
}, {
"label": "2013-01-05",
"value": "96"
}]
我寫了一個函數,從表中檢索信息,但我無法按正確的順序排列它們,也有更好的方法來做它。
function getUsersCountOnDate()
{
$result = mysql_query("Select FROM_UNIXTIME(regtime, '%Y-%m-%d') as date, count(FROM_UNIXTIME(regtime, '%Y-%m-%d')) as count from users group by FROM_UNIXTIME(regtime, '%Y-%m-%d') order by FROM_UNIXTIME(regtime, '%Y-%m-%d') DESC");
while($row = mysql_fetch_array($result)){
$date[] = $row['date'];
}
$result = mysql_query("Select FROM_UNIXTIME(regtime, '%Y-%m-%d') as date, count(FROM_UNIXTIME(regtime, '%Y-%m-%d')) as count from users group by FROM_UNIXTIME(regtime, '%Y-%m-%d') order by FROM_UNIXTIME(regtime, '%Y-%m-%d') DESC");
while($row = mysql_fetch_array($result)){
$count[] = $row['count'];
}
$merged = array_merge($date, $count);
return json_encode($merged);
}
我檢索的是這樣的:[“2016-03-18”,“2016-03-13”,“2016-03-11”,“2016-03-06”,“2016-03-04”,“6 ”, “1”, “1”, “1”, “1”]
有人可以幫我嗎?
首先,使用PDO或mysqli函數進行數據庫查詢。
array_merge()
將第一個數組放在第二個數組之前。 例如: $array1 = array('red', 'yellow'); $array2 = array('blue', 'green'); $arrayMerged = array_merge($array1, $array2); $arrayMerged is now array(red,yellow,blue,green);
$array1 = array('red', 'yellow'); $array2 = array('blue', 'green'); $arrayMerged = array_merge($array1, $array2); $arrayMerged is now array(red,yellow,blue,green);
由於您執行的查詢是相同的,您可以這樣做:
$result = mysql_query("Select FROM_UNIXTIME(regtime, '%Y-%m-%d') as date, count(FROM_UNIXTIME(regtime, '%Y-%m-%d')) as count from users group by FROM_UNIXTIME(regtime, '%Y-%m-%d') order by FROM_UNIXTIME(regtime, '%Y-%m-%d') DESC");
$i = 0;
while($row = mysql_fetch_array($result)){
$date[$i][label] = $row['date'];
$date[$i][value] = $row['count'];
$i++;
}
$newArray = json_encode($date);
嘗試
echo json_encode($merged);
自己回歸自己有時不夠。 例如,如果你想在ajax調用上接收json,則必須回顯它。
另外,你應該看看PDO而不是使用普通的舊mysql函數。
請像這樣使用它:
function getUsersCountOnDate()
{
$result = mysql_query("Select FROM_UNIXTIME(regtime, '%Y-%m-%d') as date, count(FROM_UNIXTIME(regtime, '%Y-%m-%d')) as count from users group by FROM_UNIXTIME(regtime, '%Y-%m-%d') order by FROM_UNIXTIME(regtime, '%Y-%m-%d') DESC");
while($row = mysql_fetch_array($result)){
$dates[] = $row['date'];
}
$result = mysql_query("Select FROM_UNIXTIME(regtime, '%Y-%m-%d') as date, count(FROM_UNIXTIME(regtime, '%Y-%m-%d')) as count from users group by FROM_UNIXTIME(regtime, '%Y-%m-%d') order by FROM_UNIXTIME(regtime, '%Y-%m-%d') DESC");
while($row = mysql_fetch_array($result)){
$count[] = $row['count'];
}
$dates = array("2013-01-07", "2013-01-06", "2013-01-05");
$count = array("4", "65", "96");
$i = 0;
foreach ($dates as $date)
{
$newArray[$i]['label'] = $date;
$newArray[$i]['value'] = $count[$i];
$i++;
}
///print_r($newArray);
return json_encode($newArray);
}
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