Javascript 表中的定位按钮
我正在动态地创建一个表行,如下所示,在最后一个表单元格中有一个“Delete”按钮,但在渲染时,Delete按钮不会整齐地落在表中它应该落在的位置。有什么想法吗Javascript 表中的定位按钮,javascript,jquery,html,Javascript,Jquery,Html,我正在动态地创建一个表行,如下所示,在最后一个表单元格中有一个“Delete”按钮,但在渲染时,Delete按钮不会整齐地落在表中它应该落在的位置。有什么想法吗 strAppend = '<tr>' + '<td id="SUBJ_'+idNum+'" class="td-SUBJ_'+idNum+'"><input type="text" class="SUBJ" value="Elective" name="SUB
strAppend = '<tr>' +
'<td id="SUBJ_'+idNum+'" class="td-SUBJ_'+idNum+'"><input type="text" class="SUBJ" value="Elective" name="SUBJ"> </td>' +
'<td id="REQUIRED_'+idNum+'" class="td-REQUIRED"><input type="checkbox" value="1" maxlength="40"> </td>' +
'<td id="DELETED_'+idNum+'" class="td-DELETED"><button id="DELETE_"'+idNum+'" class="DELETE">Del</button></td>' +
'<td><input value="Test"></td>' +
'</tr>'
//Add row
$j("tbody > tr:last").prev().before(strAppend);
strippend=''+
' ' +
' ' +
“tr:last”).prev()在(结束)之前;
您太早结束报价
'<td id="DELETED_'+idNum+'" class="td-DELETED"><button id="DELETE_"'+idNum+'" class="DELETE">Del</button></td>' +
'Del
去掉按钮定义中的额外引号
'<td id="DELETED_'+idNum+'" class="td-DELETED"><button id="DELETE_'+idNum+'" class="DELETE">Del</button></td>' +
“Del”+
首先从中删除id,您只能使用id一次,并更改括号“DELETE \”“+idNum+”\”。即使删除额外引号,位置仍处于关闭状态。删除了额外引号,但仍然无法解决定位问题。按钮出现,但不在“正确”位置
'<td id="DELETED_'+idNum+'" class="td-DELETED"><button id="DELETE_'+idNum+'" class="DELETE">Del</button></td>' +