Javascript 在react中将链接作为dataFormatter返回时发生不变冲突错误
在表格列标题中,我想返回导航到每个标题的编辑模式的链接。我正在使用react引导表,并在组件的构造函数中创建了一个自定义数据格式化程序Javascript 在react中将链接作为dataFormatter返回时发生不变冲突错误,javascript,reactjs,react-bootstrap-table,Javascript,Reactjs,React Bootstrap Table,在表格列标题中,我想返回导航到每个标题的编辑模式的链接。我正在使用react引导表,并在组件的构造函数中创建了一个自定义数据格式化程序 class Grid extends Component { constructor(props) { super(props); this.anchorFormatter = (cell, row, slug) => { let link = "/"+slug; return ( <Lin
class Grid extends Component {
constructor(props) {
super(props);
this.anchorFormatter = (cell, row, slug) => {
let link = "/"+slug;
return (
<Link to={link}>
{cell}
</Link>
)
}
class BlogAnchor extends Component {
constructor(props) {
super(props);
}
render() {
return (
<Link to={this.props.link}> {this.props.linkText} </Link>
);
}
}
第二部分是如何将slug的值传递给数据格式化程序?我从get request API调用中获得这样的数据
{
"title": "Experiments in DataOps",
"status": true,
"publish_date": "2020-01-29",
"slug": "experiments-in-dataops"
},
我必须使用原始的html标签,而不是链接,这样才能工作
const anchorFormat = (cell,row) => {
let link = '#/blogs/' + row.slug;
return ( <div><a className="nav-link" href={link}>{cell}</a></div>);
}
const-anchorFormat=(单元格,行)=>{
让link='#/blogs/'+row.slug;
返回();
}
另一种方法是创建一个类或函数,然后将其用作组件,这里我创建组件
class Grid extends Component {
constructor(props) {
super(props);
this.anchorFormatter = (cell, row, slug) => {
let link = "/"+slug;
return (
<Link to={link}>
{cell}
</Link>
)
}
class BlogAnchor extends Component {
constructor(props) {
super(props);
}
render() {
return (
<Link to={this.props.link}> {this.props.linkText} </Link>
);
}
}
表列引用了这个函数
<TableHeaderColumn isKey dataField="title" dataSort dataFormat {this.anchorFormatter}>Title</TableHeaderColumn>
标题
this.anchorFormatter = this.anchorFormatter.bind(this);
<TableHeaderColumn isKey dataField="title" dataSort dataFormat {this.anchorFormatter}>Title</TableHeaderColumn>