【发布时间】:2018-08-20 01:11:23
【问题描述】:
我试图在一个脚本中使用 2 个不同的表格、不同的输入字段和不同的文本。我在下面使用此脚本将这些值传递给函数参数以获取这些值。问题是我没有得到input.value,但我可以得到input1.value。我认为input1 不会进入上面的filterFunction(input)
<script type="text/javascript">
function filterFunction(table, input, total_amount_id) {
var filter, tr, td, i, totalViewable = 0;
console.log(input1.value);
console.log(input);
filter = input.value.toUpperCase();
tr = table.getElementsByTagName("tr");
for (i = 0; i < tr.length; i++) {
td = tr[i].getElementsByTagName("td")[1];
tds = tr[i].getElementsByTagName("td")[0];
if (td) {
if (td.innerHTML.toUpperCase().indexOf(filter) > -1) {
tr[i].style.display = "";
totalViewable += parseFloat(tds.innerHTML);
document.getElementById(total_amount_id).innerHTML = "$" + totalViewable.toFixed(2);
} else {
tr[i].style.display = "none";
}
}
}
}
var table1 = document.getElementById("dateTable");
var input1 = document.getElementById("event_date_range");
var total_amount_id1 = "total_amount_td";
filterFunction(table1, input1, total_amount_id1);
var table2 = document.getElementById("dateTable2");
var input2 = document.getElementById("event_date_range2");
var total_amount_id2 = "total_amount_td2";
filterFunction(table2, input2, total_amount_id2);
</script>
【问题讨论】:
标签: javascript jquery laravel datatable