Joh*_*ker 5 javascript php json extjs
当我的MySQl数据类型为bit(1)时,但是用php打印时json_encode会用unicode写吗?IIS工作正常,
但是在我的专用服务器上,Apache托管将变为unicode。为什么?

您可以看到Locating,Locating在Mysql数据类型上是位,但打印\ u0001?为什么?
这是我的编码GET方法get-googlemarker.php以获得此结果
<?php
mysql_connect("localhost", "root", "123456") or die("Could not connect");
mysql_select_db("db_gps") or die("Could not select database");
$parent_id = $_GET['mainid'];
$query = "SELECT *
FROM tbl_locate AS a
INNER JOIN
(
SELECT MainID, Max(DateTime) AS DateTime
FROM tbl_locate
GROUP BY MainID
) AS b
ON a.MainID = b.MainID
AND a.DateTime = b.DateTime
LEFT JOIN
(
SELECT b.PicData
, b.PicUploadedDateTime
, b.MainID
FROM (SELECT MainID,Max(PicUploadedDateTime) as PicUploadedDateTime
FROM tbl_picture
group by MainID
) l
JOIN tbl_picture b
ON b.MainID = l.MainID AND b.PicUploadedDateTime = l.PicUploadedDateTime
) AS c
ON a.MainID = c.MainID";
$rs = mysql_query($query);
$arr = array();
while($obj = mysql_fetch_object($rs)) {
$arr[] = $obj;
}
echo json_encode($arr);
?>
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代表性的数据是正确的,但在使用JavaScript的下面我无法读取定位值IM的问题是,我已经尝试alert了纪录,但blank.i试图定位与type:string,type:bit,type:bytes,type:int,
但不起作用。无法显示任何内容alert
Ext.define('GoogleMarkerModel', {
extend: 'Ext.data.Model',
fields: [
{name: 'ID', type: 'int'},
{name: 'Locating', type: 'int'},
{name: 'MainPower', type: 'int'},
{name: 'Acc', type: 'int'},
{name: 'PowerOff', type: 'int'},
{name: 'Alarm', type: 'int'},
{name: 'Speed', type: 'int'},
{name: 'Direction', type: 'int'},
{name: 'Latitude', type: 'float'},
{name: 'Longitude', type: 'float'},
{name: 'DateTime', type: 'datetime'},
{name: 'MainID', type: 'int'},
{name: 'IOState', type: 'int'},
{name: 'OilState', type: 'int'},
{name: 'PicUploadedDateTime', type: 'datetime'},
{name: 'PicData', type: 'str'}
]
});
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变量的表示仍然是正确的。
从 PHP 5.4 开始,您可以JSON_UNESCAPED_UNICODE使用json_encode(). 请参阅http://php.net/manual/en/function.json-encode.php
我的猜测是您的 IIS 默认返回未转义的值。
在 Javascript 中你通常会使用parseInt(\u0001)什么。因为它是extjs,所以我不知道。
计划B:
A) 将数据库列从位更改为整数。
B) 在 php 中转换数据库值
while($obj = mysql_fetch_object($rs)) {
$obj->Locating = (int)$obj-Locating;
// or try with (string) if (int) fails
$arr[] = $obj;
}
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C) 只需str_replace在 json 输出中:
$json = json_encode($arr);
$json = str_replace('"\u0001"', '"1"', $json);
$json = str_replace('"\u0000"', '"0"', $json);
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