Question

我希望找到一种从espn

中提取abcd@espn.com的方法

我在Microsoft SQL Server中工作。有谁知道怎么写substring来完成这个？

谢谢！

Answer 1

试试这个

select SUBSTRING(
  'xx.abcd@espn.com', --input string
  charindex('@', 'xx.abcd@espn.com') + 1,
  (len('xx.abcd@espn.com')-charindex('@', 'xx.abcd@espn.com'))-4
                 )

Answer 2

DECLARE @emailAddress TABLE (emailAddress VARCHAR(255))
INSERT  @emailAddress (emailAddress)
VALUES ('abcd@espn.com'),
       ('ab.cd@taylormade.edu'),
       ('notanemailaddress')

SELECT  IIF(emailAddress LIKE '%@%.%', -- Confirm it has valid structure
            SUBSTRING(emailAddress,
                      CHARINDEX('@',emailAddress)+1,
                      CHARINDEX('.',emailAddress,CHARINDEX('@',emailAddress))-(CHARINDEX('@',emailAddress)+1)), --Extract the domain
            NULL) -- Provide default value if not an emailAddress
        AS emailDomain
FROM    @emailAddress

这会考虑无效的emailAddresses值，并在遇到无效值时返回NULL（或者放在那里的任何内容）。

Answer 3

这应该有效：

<script type="text/javascript" src="d3.v3.min.js"> </script>

<script>

var width = 900,
    height = 590;

var svg = d3.select("body")
        .append("svg")
        .attr("width", width)
        .attr("height", height)

var links = [
    {source: 'H', target: 'I'},
    {source: 'H', target: 'J'},
    {source: 'I', target: 'J'},
    {source: 'J', target: 'K'},
];

var nodes = {};
links.forEach(function(link) {
    link.source = nodes[link.source] ||
        (nodes[link.source] = {name: link.source});
    link.target = nodes[link.target] ||
        (nodes[link.target] = {name: link.target});
        });

var force = d3.layout.force()
    .size([width, height])
    .nodes(d3.values(nodes))
    .links(links)
    .on('tick', tick)
    .linkDistance(100)
    .gravity(.15)
    .friction(.8)
    .linkStrength(1)
    .charge(-425)
    .chargeDistance(600)
    .start();

var link = svg.selectAll('.link')
    .data(links)
    .enter().append('line')
    .attr('class', 'link');

var node = svg.selectAll('.node')
    .data(force.nodes())
    .enter().append('circle')
    .attr('class', 'node')
    .attr('r', width * 0.01)

function tick(e) {

    node.attr('cx', function(d) { return d.x; })
        .attr('cy', function(d) { return d.y; })
        .call(force.drag);

    link.attr('x1', function(d) { return d.source.x; })
        .attr('y1', function(d) { return d.source.y; })
        .attr('x2', function(d) { return d.target.x; })
        .attr('y2', function(d) { return d.target.y; });

};

    </script>

Answer 4

你正在寻找这个。

Select substring(@email,charindex('@', @email)+1, 
charindex('.',right(@email,len(@email) - charindex('@', @email)))-1)

电子邮件地址可以在.之前和之后多次包含@，这应该会在@

之后返回域的第一部分

从EmailAddress中提取域文本

4 个答案: