Question

我有基本对象，我想搜索标题和描述值并返回找到的对象。在您的帮助之后，我可以将它们输出为HTML。我目前收到语法错误。

在该示例中，我想返回所有包含搜索词“ dog”的对象

var search_term = "dog";
var search_results = [];
var page_json = [{
  "page_title": "Display",
  "page_url": "/display/",
  "page_description": "We display stuff. search this description for the word dog.",
}, {
  "page_title": "DRAM",
  "page_url": "/dram/",
  "page_description": "stuff about dram here",
}, {
  "page_title": "TVs",
  "page_url": "/tv/",
  "page_description": "stuff about tvs here",
},];

$(page_json).each(function(i, value) {
  //console.log( value.page_description );
  // search for our term in the description AND title values
  if ($(value.page_description).contains(search_term)) {
    console.log(value);
    // put that object into search_results array
  }
});

我想得到的答案...

var search_term = "dog";
var search_results = [];
var page_json = [{
        "page_title": "Display",
        "page_url": "/display/",
        "page_description": "search this description for the word dog.",
    },{
        "page_title": "this title has a dog",
        "page_url": "/dram/",
        "page_description": "stuff about dram here",
    },{
        "page_title": "TVs",
        "page_url": "/tv/",
        "page_description": "stuff about tvs here",
    },
];

var search = function(search_term) {
    return page_json.filter(function(item) {
        // if search term is inside the title OR description
        if (item.page_description.indexOf(search_term) !== -1 || item.page_title.indexOf(search_term) !== -1){
            return item;
        }
    })
}

search_results = search(search_term);

console.log(search_results);

Answer 1

我不确定jquery是如何完成的，但是这是您在JS中的操作方式：

const foundObject = page_json.find(obj=> 
     obj.page_title.includes(search_term) ||
     obj.page_description.includes(search_term)
   )

console.log(foundObject)返回所需的输出。

如果要使用普通的for循环进行操作：

 let foundObject=null;
 for(let i=0; i < page_json.length; i++){
   const exists = page_json[i].page_title === search_term || page_json[i].page_description === search_term;
   if(exists) {
      foundObject = page_json[i]; // What you want.
   }
 }
if(foundObject) { // Do something.. }

让我知道是否不清楚！

Answer 2

您无需在此处使用Jquery。 Javascript具有过滤器功能，非常适合这种情况。

var search_term = "dog";
var search_results = [];
var page_json = [{
    "page_title": "Display",
    "page_url": "/display/",
    "page_description": "We display stuff. search this description for the word dog.",
},
{
    "page_title": "DRAM",
    "page_url": "/dram/",
    "page_description": "stuff about dram here",
},
{
    "page_title": "TVs",
    "page_url": "/tv/",
    "page_description": "stuff about tvs here",
},
];

search = function(search_term, json) {
    return json.filter(function(i) {
        if (i.page_description.indexOf('about') !== -1) return i;
    })
}

search_results = search(search_term, page_json);

console.log(search_results);

Answer 3

要搜索正确的page_title，请输入以下内容：

<?php
if (function_exists('databaseSELECTOperation')) {
    echo "function found.<br />\n";
} else {
    echo "function not found<br />\n";
}
?>

或

if(page_json[i]["page_title"] == 'whatever title' && page_json[i]["page_description"] == 'whatever description'){
//do something here
}

搜索对象以查找单词并返回结果

3 个答案: