Question

This is my code:
Let s say that name is set to john

$("#first_table").append($( #second_table ).find("tr:contains(" + name + ")"));

基本上,我在网站上有2个表格。我想首先把内容移至第二,但只有完全匹配。如何做到这一点?

Now if I have (for example) john john #1, john #2 in my table, they would all be moved. I would only like to move john and leave john #1 and john #2 Thanks! EDIT: I would like the code to work to match the variable name EXACTLY. So if name is JOHN it must only get JOHN. If it s JOHN #1, it must only get JOHN #1 and so on. Is this possible?

Answer 1

由于每一名名字均可上<代码>:内容(+名称+”),该词与所有这些内容一致。

采用<代码>:将“#”删除。

$("#first_table").append($( #second_table ).find("tr:contains(" + name + "):not(:contains( # ))"));

最新情况:

如果你想要准确了解内部的价值观。然后使用类似的东西。

$("#second_table tr td").each(function() {
   if($(this).text() ==  John ) { 
     $(this).closest("tr").appendto($("#first_table")); //find the cotaining tr and append to another table
   }
});

Answer 2

这可能是一例缺失的引述。如下文所示,在<代码>上添加单一报价。

$("#first_table").append($( #second_table ).find("tr:contains( " + name + " )"));

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