[英]Invalid JSON string
您好,我試圖為我的網站制作一個折線圖,以顯示MysQL中數據庫中的數據,但出現了錯誤的JSON字符串錯誤,沒有任何內容顯示這是我的代碼。 我以https://developers.google.com/chart/interactive/docs/php_example中的服務器端代碼為例
HTML
<html>
<head>
<!--Load the AJAX API-->
<script type="text/javascript" src="https://www.google.com/jsapi"></script>
<script type="text/javascript" src="//ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script type="text/javascript">
// Load the Visualization API and the piechart package.
google.load('visualization', '1', {'packages':['corechart']});
// Set a callback to run when the Google Visualization API is loaded.
google.setOnLoadCallback(drawChart);
function drawChart() {
var jsonData = $.ajax({
url: "getData.php",
dataType:"json",
async: false
}).responseText;
var data = new google.visualization.DataTable(); //DEFINE DATATABLE
data.addColumn('string', 'Label'); //ADD COLUMN 1
data.addColumn('number', 'Value'); //ADD COLUMN 2
console.log(jsonData);
data.addRows(jsonData); //ADD THE RECEIVED jsonData
// SET OPTIONS
var options = {'title':'REALLY PRETTY PIE CHART',
'width':400,
'height':300};
// Instantiate and draw THE chart
var chart = new google.visualization.LineChart(document.getElementById('chart_div'));
chart.draw(data, options);
}
</script>
</head>
<body>
<!--Div that will hold the pie chart-->
<div id="chart_div"></div>
</body>
</html>
PHP
<?php
include_once 'Config.php'; //configuration of my Mysql Database
$public = 'admin'; //This variable is to select the user i want
try {
$conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
$gsent = $conn->prepare("SELECT estado,Hora FROM Datos Where Usuario LIKE '$public'");
$gsent->execute();
$resultado = $gsent->fetchAll();
$resultAdoJson = json_encode($resultado);
$resulset = json_decode($resultAdoJson);
$result = array();
$i = 65;
foreach($resulset as $res) {
$result[] = array(chr($i++), intval($res->estado));
}
print json_encode($result);
}
catch (PDOException $pe) {
die("Could not connect to the database $dbname :" . $pe->getMessage());
}
?>
Mysql表只有3行,具有值的admin:
Estado Hora
這是我從php中打印得到的JSON
[{"estado":"50","0":"50","Hora":"2015-02-16","1":"2015-02-16"}, {"estado":"53","0":"53","Hora":"2015-02-16","1":"2015-02-16"},{"estado":"10","0":"10","Hora":"2015-02-16","1":"2015-02-16"}]Array
后續代碼var_dump($ resultado)
array(3) { [0]=> array(4) { ["estado"]=> string(2) "50" [0]=> string(2) "50" ["Hora"]=> string(10) "2015-02-16" [1]=> string(10) "2015-02-16" } [1]=> array(4) { ["estado"]=> string(2) "53" [0]=> string(2) "53" ["Hora"]=> string(10) "2015-02-16" [1]=> string(10) "2015-02-16" } [2]=> array(4) { ["estado"]=> string(2) "10" [0]=> string(2) "10" ["Hora"]=> string(10) "2015-02-16" [1]=> string(10) "2015-02-16" } }
我知道代碼顯示餅狀圖,但是因為我是php的新手,所以我首先不了解json或javascript的任何知識,所以我想使示例像一個。 如何將結果轉換為折線圖?
您的PHP腳本在JSON字符串旁邊輸出單詞“ Array”。 刪除行echo $resultado;
從您的PHP腳本。
編輯:此外,您還必須將結果集格式化為具有['key','value']結構的數組的數組...例如:
[
['Mushrooms', 3],
['Onions', 1],
['Olives', 1],
['Zucchini', 1],
['Pepperoni', 2]
]
根據您的情況,如下更改PHP腳本:
$resultado = $gsent->fetchAll();
$result = array();
$i = 65;
foreach($resultado as $res) {
$result[] = array(chr($i++), intval($res->estado));
}
print json_encode($result);
如您所見,我為值選擇了“ estado”,為標簽選擇了A,B,C(使用chr($i++)
)。
修改您的JavaScript:
// STORE RESPONSE OF THE AJAX REQUEST IN jsonData
var jsonString = $.ajax({
url: "getData.php",
dataType:"json",
async: false
}).responseText;
var jsonData = eval(jsonString); //create an javascript array
var data = new google.visualization.DataTable(); //DEFINE DATATABLE
data.addColumn('string', 'Label'); //ADD COLUMN 1
data.addColumn('number', 'Value'); //ADD COLUMN 2
data.addRows(jsonData); //ADD THE RECEIVED jsonData
// SET OPTIONS
var options = {'title':'REALLY PRETTY PIE CHART',
'width':400,
'height':300};
// Instantiate and draw THE chart
var chart = new google.visualization.LineChart(document.getElementById('chart_div'));
chart.draw(data, options);
結果看起來像這樣:
希望對您有所幫助。
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