javascript - 点击功能不起作用，因为它在另一个返回语句中

问题描述

我试图通过点击按钮来创造历史

li 的 onclick 功能不起作用

正如您在代码中看到的那样，<SuggestionsList/>在最后一个 return 语句中，它的函数呈现在 const SuggestionsList = (props) => { . onclick 函数在函数内部 comimg const SuggestionsList，这使得 onclick 函数不起作用

我在代码中创建了确切的工作并且它在那里工作没有任何问题我不明白为什么它在我的本地输入链接描述中不起作用

 function finddoctor(e) {
    console.log(e);
    history.push(`/detiled/${e} `);
  }

 
  const onChange = (event) => {
    const value = event.target.value;
    setInputValue(value);
    setShowResults(false);

    

    const filteredSuggestions = suggestions.filter(
      (suggestion) =>
        suggestion.firstname
          .toString()
          .toLowerCase()
          .includes(value.toLowerCase()) ||
        suggestion.id.toString().toLowerCase().includes(value.toLowerCase())
    );

  

    setFilteredSuggestions(filteredSuggestions);
    setDisplaySuggestions(true);
  };

  const onSelectSuggestion = (index) => {
    setSelectedSuggestion(index);
    setInputValue(filteredSuggestions[index]);
    setFilteredSuggestions([]);
    setDisplaySuggestions(false);
  };

  const SuggestionsList = (props) => {
  const {
      suggestions,
      inputValue,

      onSelectSuggestion,
      displaySuggestions,
      selectedSuggestion,
    } = props;

    if (inputValue && displaySuggestions) {
      if (suggestions.length > 0) {
        return (
          <ul className="suggestions-list" style={styles.ulstyle}>
            {suggestions.map((suggestion, index) => {
    
              const isSelected = selectedSuggestion === index;
              const classname = `suggestion ${isSelected ? "selected" : ""}`;
              return (
                <>
              
                  <li
                    style={styles.listyle}
                    onClick={finddoctor(suggestion.id)}
                    key={index}
                    className={classname}
                  >
                    {suggestion.firstname}
                  </li>
                </>
              );
            })}
          </ul>
        );
      } else {
        return <div>No suggestions available...</div>;
      }
    }
    return <></>;
  };


  useEffect(() => {
    axios
      .get("admin-panel/all-doctors-list/")
      .then((res) => {
        const data = res.data;
      setShowSerch(data);

      });
  
  }, []);

  return (
    <>
      <div className="note-container" style={styles.card}>
        <div style={styles.inner}>
          <p style={{ textAlign: "left" }}>Search Doctors</p>
          <form className="search-form" style={{}}>
            {showResults ? (
              <FontAwesomeIcon
                style={{ marginRight: "-23px" }}
                icon={faSearch}
              />
            ) : null}
            <input
              onChange={onChange}
              value={inputValue}
              style={styles.input}
              type="Search"
            />

            <SuggestionsList
              inputValue={inputValue}
              selectedSuggestion={selectedSuggestion}
              onSelectSuggestion={onSelectSuggestion}
              displaySuggestions={displaySuggestions}
              suggestions={filteredSuggestions}
            />
          </form>
         
        </div>
      </div>
    </>
  );
};

标签： javascriptreactjs

javascript - 点击功能不起作用，因为它在另一个返回语句中

问题描述

解决方案

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