我有两个html表,它们动态地使用不同的数据。无法将两者导出到Excel工作表我愿意导出单个表。
<div class="data" style="display: block;">
<table id="testTable" border="1">
<tbody>
<table style="width:300px">
<tr>
<td>John</td>
<td>Doe</td>
<td>80</td>
</tr>
<tbody>
</table>
</div>
<div class="data" style="display: block;">
<table id="testTable" border="1">
<tbody>
<table style="width:300px">
<tr>
<td>vsr</td>
<td>sub</td>
<td>90</td>
</tr>
<tbody>
</table>
</div>
像这样,我的表带有来自db
的不同数据集我的js代码是
var tableToExcel = (function() {
var uri = 'data:application/vnd.ms-excel;base64,'
, template = '<html xmlns:o="urn:schemas-microsoft-com:office:office" xmlns:x="urn:schemas-microsoft-com:office:excel" xmlns="http://www.w3.org/TR/REC-html40"><head><!--[if gte mso 9]><xml><x:ExcelWorkbook><x:ExcelWorksheets><x:ExcelWorksheet><x:Name>{worksheet}</x:Name><x:WorksheetOptions><x:DisplayGridlines/></x:WorksheetOptions></x:ExcelWorksheet></x:ExcelWorksheets></x:ExcelWorkbook></xml><![endif]--></head><body><table>{table}</table></body></html>'
, base64 = function(s) { return window.btoa(unescape((s))) }
, format = function(s, c) { return s.replace(/{(\w+)}/g, function(m, p) { return c[p]; }) }
return function(table, name, filename) {
if (!table.nodeType) table = document.getElementById(resultarea)
var ctx = {worksheet: name || 'Worksheet', table: resultarea.innerHTML}
document.getElementById("dlink").href = uri + base64(format(template, ctx));
document.getElementById("dlink").download = filename;
document.getElementById("dlink").click();
}
})()
无法同时导出到excel两个表数据。请帮帮我。
答案 0 :(得分:0)
您应该通过classname引用,而不是通过id引用。对于您现在遇到的错误类型,建议不要对多个HTML元素使用相同的ID。
答案 1 :(得分:0)
done = False
while not done:
print('=========================================')
print(' Aria\'s Pet Store Point of Sale System')
print('=========================================')
print('[1] Add Products to Cart')
print('[2] Remove Products to Cart')
print('[3] View Cart')
print('[4] Generate Reciept')
print('[5] Logout')
print('[6] Exit')
print('=========================================')
choice = input('Choice: ')
if choice == '1':
pass
elif choice == '2':
pass
elif choice == '3':
pass
elif choice == '4':
pass
elif choice == '5':
pass
elif choice == '6':
exit(0)
else:
print('Invalid Input')