无法在 parseJSON 中检测到 "unexpected identifier"
unable to detect "unexpected identifier" in parseJSON
我正在尝试从数据库中获取数据并执行 json_encode
$data = json_encode($results);
当我输入 echo($data);死();我得到了以下结果
[{"CAL_DATE":"01-JUN-13","CUSTOMER_TEXT":"Wa","PRODUCT_GROUP":"ACP'S","NET_SALES":"187002.04","RANK":"1"},{"CAL_DATE":"01-JUN-13","CUSTOMER_TEXT":"CVS","PRODUCT_GROUP":"ACP'S","NET_SALES":"127948.68","RANK":"2"},{"CAL_DATE":"01-JUN-13","CUSTOMER_TEXT":"McK","PRODUCT_GROUP":"ACP'S","NET_SALES":"81079.29","RANK":"3"},{"CAL_DATE":"01-JUN-13","CUSTOMER_TEXT":"Car","PRODUCT_GROUP":"ACP'S","NET_SALES":"65320.42","RANK":"4"},{"CAL_DATE":"01-JUN-13","CUSTOMER_TEXT":"Krog.","PRODUCT_GROUP":"ACP'S","NET_SALES":"31977.95","RANK":"5"}]
在 jquery 代码中,我尝试按如下方式解析 JSON $data
$(function () {
var data = new Array();
data = $.parseJSON('<?php echo $data; ?>'); //error occuring here
//other code goes here
});
我在 data = $.parseJSON('<?php echo $data; ?>');
附近收到错误 Uncaught SyntaxError: Unexpected identifier
我在
中得到以下输出
[{"CAL_DATE":"01-JUN-13","CUSTOMER_TEXT":"Wa","PRODUCT_GROUP":"ACP'S","NET_SALES":"187002.04","RANK":"1"},{"CAL_DATE":"01-JUN-13","CUSTOMER_TEXT":"CVS,Inc.","PRODUCT_GROUP'S":"ACP","NET_SALES":"127948.68","RANK":"2"},{"CAL_DATE":"01-JUN-13","CUSTOMER_TEXT":"McK","PRODUCT_GROUP":"ACP'S","NET_SALES":"81079.29","RANK":"3"},{"CAL_DATE":"01-JUN-13","CUSTOMER_TEXT":"Car","PRODUCT_GROUP":"ACP'S","NET_SALES":"65320.42","RANK":"4"},{"CAL_DATE":"01-JUN-13","CUSTOMER_TEXT":"Krog.","PRODUCT_GROUP":"ACP'S","NET_SALES":"31977.95","RANK":"5"}]
任何人都可以告诉我为什么我在该行中遇到错误吗?提前致谢
这是一个有效的 json,因此您不需要解析它:
data = <?php echo $data; ?>
我猜你想这样使用它:
$(function () {
var data = '<?php echo $data; ?>';
data = $.parseJSON(data); //error occuring here
//other code goes here
});
来自文档:
Takes a well-formed JSON string and returns the resulting JavaScript value.
从文档中查看:
var obj = jQuery.parseJSON( '{ "name": "John" }' ); // results in {"name" : "John"}
alert( obj.name === "John" );
'{ "name": "John" }'
是 json 格式的字符串。
我正在尝试从数据库中获取数据并执行 json_encode
$data = json_encode($results);
当我输入 echo($data);死();我得到了以下结果
[{"CAL_DATE":"01-JUN-13","CUSTOMER_TEXT":"Wa","PRODUCT_GROUP":"ACP'S","NET_SALES":"187002.04","RANK":"1"},{"CAL_DATE":"01-JUN-13","CUSTOMER_TEXT":"CVS","PRODUCT_GROUP":"ACP'S","NET_SALES":"127948.68","RANK":"2"},{"CAL_DATE":"01-JUN-13","CUSTOMER_TEXT":"McK","PRODUCT_GROUP":"ACP'S","NET_SALES":"81079.29","RANK":"3"},{"CAL_DATE":"01-JUN-13","CUSTOMER_TEXT":"Car","PRODUCT_GROUP":"ACP'S","NET_SALES":"65320.42","RANK":"4"},{"CAL_DATE":"01-JUN-13","CUSTOMER_TEXT":"Krog.","PRODUCT_GROUP":"ACP'S","NET_SALES":"31977.95","RANK":"5"}]
在 jquery 代码中,我尝试按如下方式解析 JSON $data
$(function () {
var data = new Array();
data = $.parseJSON('<?php echo $data; ?>'); //error occuring here
//other code goes here
});
我在 data = $.parseJSON('<?php echo $data; ?>');
Uncaught SyntaxError: Unexpected identifier
我在
中得到以下输出 [{"CAL_DATE":"01-JUN-13","CUSTOMER_TEXT":"Wa","PRODUCT_GROUP":"ACP'S","NET_SALES":"187002.04","RANK":"1"},{"CAL_DATE":"01-JUN-13","CUSTOMER_TEXT":"CVS,Inc.","PRODUCT_GROUP'S":"ACP","NET_SALES":"127948.68","RANK":"2"},{"CAL_DATE":"01-JUN-13","CUSTOMER_TEXT":"McK","PRODUCT_GROUP":"ACP'S","NET_SALES":"81079.29","RANK":"3"},{"CAL_DATE":"01-JUN-13","CUSTOMER_TEXT":"Car","PRODUCT_GROUP":"ACP'S","NET_SALES":"65320.42","RANK":"4"},{"CAL_DATE":"01-JUN-13","CUSTOMER_TEXT":"Krog.","PRODUCT_GROUP":"ACP'S","NET_SALES":"31977.95","RANK":"5"}]
任何人都可以告诉我为什么我在该行中遇到错误吗?提前致谢
这是一个有效的 json,因此您不需要解析它:
data = <?php echo $data; ?>
我猜你想这样使用它:
$(function () {
var data = '<?php echo $data; ?>';
data = $.parseJSON(data); //error occuring here
//other code goes here
});
来自文档:
Takes a well-formed JSON string and returns the resulting JavaScript value.
从文档中查看:
var obj = jQuery.parseJSON( '{ "name": "John" }' ); // results in {"name" : "John"}
alert( obj.name === "John" );
'{ "name": "John" }'
是 json 格式的字符串。