我正在使用PHP创建一些似乎无效的JSON数据。我正在尝试将google API集成到我的代码中。
<?php
$con=mysql_connect("localhost","root","") or die("Failed to connect with database!!!!");
mysql_select_db("mobiledb", $con);
// The Chart table contains two fields: weekly_task and percentage
// This example will display a pie chart. If you need other charts such as a Bar chart, you will need to modify the code a little to make it work with bar chart and other charts
$sth = mysql_query("SELECT `id`, `Q1`, `Q2` FROM `table2` WHERE `id`=8710058770");
/*
---------------------------
example data: Table (Chart)
--------------------------
weekly_task percentage marks
Sleep 30 60
Watching Movie 40 80
work 44 90
*/
$rows = array();
//flag is not needed
$flag = true;
$table = array();
$table['cols'] = array(
// Labels for your chart, these represent the column titles
// Note that one column is in "string" format and another one is in "number" format as pie chart only required "numbers" for calculating percentage and string will be used for column title
array('label' => 'id', 'type' => 'string'),
array('label' => 'Q1', 'type' => 'number'),
array('label' => 'Q2', 'type' => 'number')
);
$rows = array();
while($r = mysql_fetch_assoc($sth)) {
$temp = array();
// the following line will be used to slice the Pie chart
$temp[] = array('v' => (string) $r['id']);
// Values of each slice
$temp[] = array('v' => (int) $r['Q1'], 'f' => (int) $r['Q2']);
$rows[] = array('c' => $temp);
}
$table['rows'] = $rows;
$jsonTable = json_encode($table);
echo $jsonTable;
?>
MY Json输出并检入http://jsonlint.com/
{
cols: [
{
label: "id",
type: "string"
},
{
label: "Q1",
type: "number"
},
{
label: "Q2",
type: "number"
}
],
rows: [
{
c: [
{
v: "8710058770"
},
{
v: 35,
f: 40
}
]
},
{
c: [
{
v: "8710058770"
},
{
v: 60,
f: 70
}
]
},
{
c: [
{
v: "8710058770"
},
{
v: 75,
f: 85
}
]
}
]
}
我验证时可以看到的错误
第1行的解析错误：
{cols：[{
----- ^
期待'STRING'，'}'
print_r（$ table）output
Array
(
[cols] => Array
(
[0] => Array
(
[label] => id
[type] => string
)
[1] => Array
(
[label] => Q1
[type] => number
)
[2] => Array
(
[label] => Q2
[type] => number
)
)
[rows] => Array
(
[0] => Array
(
[c] => Array
(
[0] => Array
(
[v] => 8710058770
)
[1] => Array
(
[v] => 35
)
[2] => Array
(
[v] => 40
)
)
)
[1] => Array
(
[c] => Array
(
[0] => Array
(
[v] => 8710058770
)
[1] => Array
(
[v] => 60
)
[2] => Array
(
[v] => 70
)
)
)
[2] => Array
(
[c] => Array
(
[0] => Array
(
[v] => 8710058770
)
[1] => Array
(
[v] => 75
)
[2] => Array
(
[v] => 85
)
)
)
)
)
{"cols":[{"label":"id","type":"string"},{"label":"Q1","type":"number"},{"label":"Q2","type":"number"}],"rows":[{"c":[{"v":"8710058770"},{"v":35},{"v":40}]},{"c":[{"v":"8710058770"},{"v":60},{"v":70}]},{"c":[{"v":"8710058770"},{"v":75},{"v":85}]}]}
答案 0 :(得分：4)
在<?php之前，你的php文件开头有一些空格，这会导致输出被错过解释。
header("Content-Type: application/json");
。
答案 1 :(得分：0)
实际上问题是由于字符串值没有引号。 结果应该是
{
"cols": [{
"label": "id",
"type": "string"
},
{
"label": "Q1",
"type": "number"
},
{
"label": "Q2",
"type": "number"
}],
"rows": [{
"c": [{
"v": "8710058770"
},
{
"v": 35,
"f": 40
}]
},
{
"c": [{
"v": "8710058770"
},
{
"v": 60,
"f": 70
}]
},
{
"c": [{
"v": "8710058770"
},
{
"v": 75,
"f": 85
}]
}]
}
因为你的密钥没有用引号括起来，所以这是错误的。
现在的问题是为什么它不应该在引号中。