在coldfusion中有json_encode(PHP)的替代品吗?
我使用ColdFusion在coldfusion中有json_encode(PHP)的替代品吗?,php,json,coldfusion,Php,Json,Coldfusion,我使用ColdFusionSerializeJSON(queryname)将我的查询结果转换为json格式,结果如下: { "COLUMNS":["ID","TAG","TAG_DESCRIPTION","TYPE"] , "DATA": [1,"PHP","PHP is an open-source server-side scripting language widely used in web development.","programming"] } 但是我想要json这种
SerializeJSON(queryname)
将我的查询结果转换为json格式,结果如下:{
"COLUMNS":["ID","TAG","TAG_DESCRIPTION","TYPE"]
, "DATA": [1,"PHP","PHP is an open-source server-side scripting language widely used in web development.","programming"]
}
但是我想要json这种格式,这是PHPjson\u encode($rows)
返回的:在ColdFusion中是否有
json_encode(PHP)
的替代品?或者我必须手动创建这个json格式吗
谢谢不,CFML中没有函数以PHP决定用于表示其内部数据类型之一的JSON序列化的格式创建JSON数据包,正如PHP中没有函数以Adobe任意决定的格式创建JSON,将其内部数据类型之一表示为JSON一样 你得自己滚 JSON
JSON\u encode()
吐出来的东西有点乱。用数字索引和字符串键将键/值加倍是怎么回事?根据你给我们的,我不太清楚第二行是如何表示的。你能给我们几行编码为JSON的相当通用的数据,让我们看看PHP在做什么吗
我确信这只是一个循环您的CFML记录集的问题,然后为每一行的列值填充数字/字符串键控属性。不要试图手工构建JSON,在正确的模式中构建本机CFML结构(或结构数组?),然后使用serializeJson()
将其序列化
抱歉说得有点含糊。如果您可以通过按选择的方式序列化记录来解释PHP的想法,那么我可以更好地解释CFML逻辑以达到相同的目的。您可以检查我的jQuery插件以解析本机CF查询返回。您可以对CFC进行本机远程调用,在ColdFusion的本机JSON解析中获取数据(在ajax请求中使用returnFormat:“JSON”选项),然后通过该插件传递结果,以获得您在许多其他jQuery(和其他)插件中所期望的标准名称/值格式。我也有一个类似的ExtJS数据存储解析器 如果您有CF11:
<cfcomponent>
<cfprocessingDirective pageencoding="utf-8">
<cfset this.name = "Curso">
<cfset this.serialization.preservecaseforstructkey = true >
<cfset this.serialization.serializeQueryAs = "struct">
</cfcomponent>
您将在cfm/cfc中使用serializeJSON(查询)。或:
<cffunction name="estructura" access="remote" returntype="any" >
<cfargument name="query" type="query" required="true" >
<cfset columns = arguments.query.ColumnList>
<cfset array = ArrayNew( 1 )>
<cfloop query="#arguments.query#">
<cfset query_ = structNew()>
<cfloop list="#columns#" index="col">
<cfset query_[lcase(col)] = arguments.query[col][arguments.query.CurrentRow]>
</cfloop>
<cfset array[arguments.query.CurrentRow] = query_>
</cfloop>
<cfset data.data = array>
<cfreturn data>
</cffunction>
<cffunction name="prueba" access="remote" returntype="any" returnformat="JSON">
<cfquery name="query">
select * from rsosa.colaboradores
</cfquery>
<cfreturn estructura(query)>
</cffunction>
从rsosa.colaboradores中选择*
只有在您首先选择的情况下,才能获得以下结果。
{"COLUMNS":["ID","NAME"],"DATA":[[1,"ABC"],[2,"Imperial"]]}
<!---Instead of using SerializeJSON(getQuery) to Change the getQuery result to json i am using the below way.--->
<cfset var aTmp = arraynew(1)>
<cfif #getQuery.recordcount# gt 0>
<cfloop query="getQuery">
<cfset coInfo = StructNew()>
<cfset coInfo["id"]="#getQuery.field1#">
<cfset coInfo["name"]="#getQuery.field2#">
<cfset arrayAppend(aTmp,coInfo)>
</cfloop>
<cfset jsondata = SerializeJSON(aTmp)>
<cfoutput>#jsondata#</cfoutput>
</cfif>
[{"name":"ABC","id":1},{"name":"Imperial","id":2}]
<cfcomponent>
<cfprocessingDirective pageencoding="utf-8">
<cfset this.name = "Curso">
<cfset this.serialization.preservecaseforstructkey = true >
<cfset this.serialization.serializeQueryAs = "struct">
</cfcomponent>
<cffunction name="estructura" access="remote" returntype="any" >
<cfargument name="query" type="query" required="true" >
<cfset columns = arguments.query.ColumnList>
<cfset array = ArrayNew( 1 )>
<cfloop query="#arguments.query#">
<cfset query_ = structNew()>
<cfloop list="#columns#" index="col">
<cfset query_[lcase(col)] = arguments.query[col][arguments.query.CurrentRow]>
</cfloop>
<cfset array[arguments.query.CurrentRow] = query_>
</cfloop>
<cfset data.data = array>
<cfreturn data>
</cffunction>
<cffunction name="prueba" access="remote" returntype="any" returnformat="JSON">
<cfquery name="query">
select * from rsosa.colaboradores
</cfquery>
<cfreturn estructura(query)>
</cffunction>