我有下面的代码,我试图用我保持对JSON解析器收到错误JSON.parse解析具有HTML内容
var data = JSON.parse('[{"thisFieldname":"item-company-1","thisFieldHTML":"\n\t\t\t\t\t<div class=\"new-company-field field-item\">\n\t\t\t\t\t\t<div class=\"fake-data\">\n\t\t\t\t\t\t\tCompany\n\t\t\t\t\t\t</div>\n\t\t\t\t\t</div>\n\t\t\t\t<div class=\"ui-resizable-handle ui-resizable-e\" style=\"z-index: 90; display: block;\"></div><div class=\"ui-resizable-handle ui-resizable-s\" style=\"z-index: 90; display: block;\"></div><div class=\"ui-resizable-handle ui-resizable-se ui-icon ui-icon-gripsmall-diagonal-se\" style=\"z-index: 90; display: block;\"></div>","dataFieldName":"item-company-1","locationIndex":"0","locationLeft":"427.891px","locationTop":"88.5625px","itemWidth":"100px","itemHeight":"34px","fieldRole":"","fieldDefault":"","fieldTooltip":"","fieldValidationRule":"","fieldValidationCharSet":"","fieldValidationDateFormat":"","fieldDisplayFormat":"","fieldValidationCountry":"","fieldValidationMaxLen":"","fieldValidationMinVal":"","fieldValidationMaxVal":"","fieldValidationRegExp":"","fieldValidationFormula":"","fieldValidationErrMsg":"","valid":"","condition-field":"","condition-type":"","condition-value-select":"","fontName":"","fontSize":"","fontAlign":"","fieldColorPicker":"","fieldRequired":"false","fieldReadOnly":"false","fieldMasked":"false","fieldMultiline":"false"}]');
的JSON被认为是JSON时抛出错误我试过时有效的JSON https://jsonformatter.curiousconcept.com/
什么是你的JSON代码的源代码?如果JSON是硬编码的,那么你应该直接把它分配给'data'变量而不用调用'JSON.parse()'。 –
从服务器发送JSON并将其存储在字段中作为值,以便它可以通过javascript访问。这是我可以考虑将JSON直接从PHP传递到Javascript的唯一方法,可以通过 – eqiz