在我工作的jqgrid中,我根据第一行中的数据将格式化程序应用于多个列。因此,如果第一行包含内容值" PERCENT",我将一个formtter应用于实际值列,该值将值格式化为带小数位的数字。
我遇到的问题是,当列包含数值或null时,格式化程序会将空值格式化为" 0.00"。
我设置了以下jsFiddle以显示正在发生的事情。我需要的是var1Value,var2Value和var3Value列显示空白,而不是" 0.00",当json值为null时。这可能吗?
https://jsfiddle.net/msobczak/7prbs3tu/6/
代码如下:
var mainGrid = {
"total": 1,
"page": 1,
"pageSize": 20,
"records": 1,
"rows": [{
"id": 7259,
"var1Name": "2015 Median Age",
"var1Contents": "MEDIAN",
"var1IsString": 0,
"var1IsNumber": 1,
"var1Value": "44",
"var2Name": "% '15 HHs",
"var2Contents": "PERCENT",
"var2IsString": 0,
"var2IsNumber": 1,
"var2Value": "2.07",
"var3Name": "Wine At Home",
"var3Contents": "INDEX",
"var3IsString": 0,
"var3IsNumber": 1,
"var3Value": "135"
}, {
"id": 7259,
"var1Name": "2015 Median Age",
"var1Contents": "MEDIAN",
"var1IsString": 0,
"var1IsNumber": 1,
"var1Value": null,
"var2Name": "% '15 HHs",
"var2Contents": "PERCENT",
"var2IsString": 0,
"var2IsNumber": 1,
"var2Value": null,
"var3Name": "Wine At Home",
"var3Contents": "INDEX",
"var3IsString": 0,
"var3IsNumber": 1,
"var3Value": null
}]
};
GridFunctions = {
formatVariable: function(gridId, columnName, variableValue, variableContents, isNumber) {
if (variableValue != undefined && variableContents != undefined) {
switch (variableContents) {
case "MEDIAN":
$(gridId).jqGrid("setColProp", columnName, {
formatter: 'number',
defaultvalue: null
});
$(gridId).jqGrid("setColProp", columnName, {
formatoptions: {
thousandsSeparator: ',',
decimalPlaces: 0
}
});
break;
case "COUNT":
$(gridId).jqGrid("setColProp", columnName, {
formatter: 'number'
});
$(gridId).jqGrid("setColProp", columnName, {
formatoptions: {
thousandsSeparator: ',',
decimalPlaces: 0
}
});
break;
case "RATIO":
$(gridId).jqGrid("setColProp", columnName, {
formatter: 'number'
});
$(gridId).jqGrid("setColProp", columnName, {
formatoptions: {
thousandsSeparator: ',',
decimalPlaces: 0
}
});
break;
case "PERCENT":
$(gridId).jqGrid("setColProp", columnName, {
formatter: 'number',
defaultvalue: null
});
$(gridId).jqGrid("setColProp", columnName, {
formatoptions: {
thousandsSeparator: ',',
decimalPlaces: 2
}
});
break;
case "INDEX":
if (isNumber == 1) {
$(gridId).jqGrid("setColProp", columnName, {
formatter: 'number'
});
$(gridId).jqGrid("setColProp", columnName, {
formatoptions: {
thousandsSeparator: ',',
decimalPlaces: 0
}
});
}
break;
}
}
}
}
$(document).ready(function() {
$("#jqGrid").jqGrid({
datatype: function(postdata) {
$('#' + 'load_' + 'jqGrid').show();
var json = mainGrid;
var thisGridId = "#jqGrid";
var columnName = "var1Value";
var varName = json.rows[0].var1Name;
// Dynamically change column header for the variable 1 column
GridFunctions.formatVariable(thisGridId, columnName, json.rows[0].var1Value, json.rows[0].var1Contents, json.rows[0].var1IsNumber);
varName = json.rows[0].var2Name;
columnName = "var2Value";
// Dynamically change column header for the variable 2 column
GridFunctions.formatVariable(thisGridId, columnName, json.rows[0].var2Value, json.rows[0].var2Contents, json.rows[0].var2IsNumber);
varName = json.rows[0].var3Name;
columnName = "var3Value";
// Dynamically change column header for the variable 3 column
GridFunctions.formatVariable(thisGridId, columnName, json.rows[0].var3Value, json.rows[0].var3Contents, json.rows[0].var3IsNumber);
var thegrid = $('#jqGrid')[0];
thegrid.addJSONData(json);
$('#' + 'load_' + 'jqGrid').hide();
},
page: 1,
colModel: [{
label: 'ID',
name: 'id',
width: 50,
hidden: false,
key: true,
editable: true,
sortable: false,
editrules: {
edithidden: true
}
},
// Variable 1
{
label: 'var1Value',
name: 'var1Value',
width: 90,
sortable: true,
hidden: false,
align: 'right'
},
// Variable 2
{
label: 'var2Value',
name: 'var2Value',
width: 90,
sortable: true,
hidden: false,
align: 'right'
},
// Variable 3
{
label: 'var3Value',
name: 'var3Value',
width: 90,
sortable: true,
hidden: false,
align: 'right'
}
],
viewrecords: true,
width: 800,
shrinkToFit: false,
height: '100%',
rowNum: 20,
pager: "#jqGridPager"
});
});
答案 0 :(得分:2)
首先,您的代码很难阅读。如果我正确理解您的问题,那么您只需使用formatter: "number"
并设置默认值。
您当前的代码使用
$(gridId).jqGrid("setColProp", columnName, {
formatter: 'number',
defaultvalue: null
});
$(gridId).jqGrid("setColProp", columnName, {
formatoptions: {
thousandsSeparator: ',',
decimalPlaces: 0
}
});
与
相同$(gridId).jqGrid("setColProp", columnName, {
formatter: 'number',
defaultvalue: null,
formatoptions: {
thousandsSeparator: ',',
decimalPlaces: 0
}
});
代码包含一些错误。您需要的属性的正确名称是defaultValue
而不是defaultvalue
,并且需要在formatoptions
内设置属性。如果您需要显示空单元格,则可以使用defaultValue: ""
或更好defaultValue: " "
:
$(gridId).jqGrid("setColProp", columnName, {
formatter: 'number',
formatoptions: {
thousandsSeparator: ',',
decimalPlaces: 0,
defaultValue: " "
}
});
此外,我不建议您使用datatype
作为功能。您将无任何优势地禁用jqGrid的许多有用功能。在我看来,您的真实代码从服务器加载数据,并且您希望根据服务器响应中的数据更改列属性。您可以在案例中使用datatype: "json"
并使用beforeProcessing
回调来预处理"数据之前的数据将由jqGrid处理。有关详细信息,请参阅the answer。