我正在尝试显示从jquery post收到的数据。但它给了我一个错误,说未定义
我的php-json编码数组是这样的。
[{"matId":"7","matName":"test","matBrand":"est","matPackaging":"1","matWidth":"434","matHeight":"23","matLength":"23","matWeight":"23","matArea":"23","matVolume":"23","matPerPack":"32","supplier1":"19","supplier2":"19","supplier3":"19","requiredInPhase":"","stockItem":"1"}]
然后在jquery中,我试图提醒其中一个值,但它给出了错误"undefined"。
这是我的jquery代码
$('#matId').on('change', function() {
var matId = $(this).val();
$.post('<?php echo base_url() . 'display_mat_details'; ?>', {matId: matId}, function(redata) {
var obj = $.parseJSON(redata);
alert(obj.matId);
});
});
您的数据包含在作为数组的[]
中。因此,您需要使用[0]
从阵列中获取第一个对象
$('#matId').on('change', function() {
var matId = $(this).val();
$.post('<?php echo base_url() . '
display_mat_details '; ?>', {
matId: matId
}, function(redata) {
var obj = $.parseJSON(redata);
alert(obj[0].matId);
});
});
示例:
var str = [{"matId":"7","matName":"test","matBrand":"est","matPackaging":"1","matWidth":"434","matHeight":"23","matLength":"23","matWeight":"23","matArea":"23","matVolume":"23","matPerPack":"32","supplier1":"19","supplier2":"19","supplier3":"19","requiredInPhase":"","stockItem":"1"}];
alert(str[0].matId);
这是因为JSON是一个包含对象的数组。。。删除JSON周围的方括号。
并使用
$.parseJSON()
你的阵列结构应该像这个
var Content = '{"matId":"7","matName":"test","matBrand":"est","matPackaging":"1","matWidth":"434","matHeight":"23","matLength":"23","matWeight":"23","matArea":"23","matVolume":"23","matPerPack":"32","supplier1":"19","supplier2":"19","supplier3":"19","requiredInPhase":"","stockItem":"1"}';
var obj = $.parseJSON(Content);
alert(obj.matId);
使用此代码用obj[0].matId
代替obj.matId
$('#matId').on('change', function() {
var matId = $(this).val();
$.post('<?php echo base_url() . 'display_mat_details'; ?>', {matId: matId}, function(redata) {
// var obj = $.parseJSON(redata);// error here structure of redata is wrong
// alert(obj[0].matId);
alert(redata[0].matId);
});
});