仅包含()在String的第一个单词中

时间:2018-06-12 09:42:44

标签: javascript string methods

您好我只想在String的第一个单词上运行includes()方法。我发现有一个可选参数fromIndex。我宁愿指定toIndex哪个值是第一个空格的索引,但似乎不存在。

你知道我怎么能做到这一点吗?谢谢!

4 个答案:

答案 0 :(得分:0)

你说过你有ngAfterViewInit(): void { jQuery('.ui-tree-toggler').addClass('d-none'); jQuery('.ui-treenode-content').addClass('shadow'); jQuery('.ui-treenode-content').addClass('fp-tree-node'); this.nodes.forEach(node => { this.setClasses(node); }); } setClasses(node: TreeNode) { if(!node.children || node.children.length === 0) { let thisNode = jQuery(`#${node.data.id}`); let styleParent = thisNode.closest(".ui-treenode") styleParent[0].classList.add("pl-0"); } if(!node.data.parentId) { let thisNode = jQuery(`#${node.data.id}`); let styleParent = thisNode.closest(".ui-treenode") styleParent[0].classList.add("pr-0"); } if (node.children && node.children.length > 0) { node.children.forEach(child => { this.setClasses(child); }); } } ,所以有两个选择:

  1. 改为使用toIndex

    indexOf
  2. 拆分第一个单词(使用var n = str.indexOf(substr); if (n != -1 && n < toIndex) { // It's before `toIndex` } split或其他),然后使用substring

    includes
  3. (当然,根据您是想要包含还是排他,调整以上if (str.substring(0, toIndex).includes(substr)) { // It's before `toIndex` } 的使用。)

答案 1 :(得分:0)

如果它是一个句子,只需将字符串拆分以获得第一个单词

&#13;
&#13;
myString = "This is my string";
firstWord = myString.split(" ")[0];
console.log("this doesn't include what I'm looking for".includes(firstWord));
console.log("This does".includes(firstWord));
&#13;
&#13;
&#13;

答案 2 :(得分:0)

您可以尝试以下方法并创建新方法

&#13;
&#13;
let str = "abc abdf abcd";
String.prototype.includes2 = function(pattern,from =0,to = 0) {
  to = this.indexOf(' ');
  to = to >0?to: this.length();
  return this.substring(from, to).includes(pattern);
}
console.log(str.includes2("abc",0,3));
console.log(str.includes2("abc",4,8));
console.log(str.includes2("abc"));
console.log(str.includes2("abd"))
&#13;
&#13;
&#13;

答案 3 :(得分:0)

您可以拆分字符串并传递给索引为0的include方法

&#13;
&#13;
var a = "this is first word of a String";
console.log(a.includes(a.split(' ')[0]));
&#13;
&#13;
&#13;