按列名显示搜索结果

Question

我有一个搜索输入，它会在您键入时从我的数据库中提取前 10 个结果，并即时消除那些不匹配的结果。我从 https://codeforgeek.com/2014/09/ajax-search-box-php-mysql/ 得到这个例子，它使用 Twitter typeahead JavaScript 库。

我的问题是，我必须定义要搜索的列，而不是在 process-search.php 中选择它。我需要的是从选择字段中选择一个表列，然后使用所选列填充搜索。我觉得我很接近，但到目前为止还没有奏效。

我已经在有效的查询和代码下方注释掉了我尝试使用的查询和代码。 Alos 我已经把选择框​​留在了 html 中。

任何帮助将非常感谢。

Index.php

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script type="text/javascript" src="../inc/js/typeahead.min.js"></script>
<script type="text/javascript" src="search-script.js"></script>

<div id="search-topic-group" class="dhl-group">
    <label for="term">Search By:</label>
    <select name="term">
        <option value="">-- Select a Search Term --</option>
        <option value="username">Username</option>
        <option value="city">City</option>
    </select>
</div>

<div id="group" class="dhl-group">
  <label for="keyword">Enter a Keyword:</label>
  <input type="text" name="keyword" class="typeahead tt-query" autocomplete="off" spellcheck="false" placeholder="Type your Query">
</div>

search-script.js

// JavaScript Document
var col = 'country'
$(document).ready(function(){
    "use strict";
     //Type Ahead Functions for the Search
 $('select[name="term"]').change(function(){
     col = $('select[name="term"]').val();
});
$('input.typeahead').typeahead({
    name: 'keyword',
    remote:'process-search.php?key=%QUERY&col=' + col,
    limit : 10
  });
});

process-search.php

<?php
    $key=$_GET['key'];
    $col = $_GET['col']
    //$term = $_GET['term'];

    $array = array();
    $errors = array();
    $db = mysqli_connect(<HOST>,<UID>,<PWD>,<DB>);
    if($db->connect_errno > 0){
        die('Unable to connect to database [' . $db->connect_error . ']');
    }
    $sql = "select * from table_name where ".$col." LIKE '%{$key}%'";

    if(!$result = $db->query($sql)){
        die('There was an error running the query [' . $db->error . ']');
    }
    while($row = $result->fetch_assoc()){
        $array[] = $row[$col];
    }
    echo json_encode($array);
?>

Answer 1

有点难看，但这应该可以工作

jquery：

var col = "default"; //Change it to whatever you want your column to default to.

$(document).ready(function(){
    "use strict";
    //Type Ahead Functions for the Search

    $('input.typeahead').typeahead({
        name: 'keyword',
        remote:'process-search.php?key=%QUERY&col='+col,
        limit : 10
      });

   //Grabs the currently selected column when it changes
   $("select[name='term']").change(function(){
      col = $("select[name='term']").val();
   })
 });

过程搜索.php

<?php
    $key=$_GET['key'];
    $col=$_GET['col'];

    $array = array();
    $errors = array();
    $db = mysqli_connect(<HOST>,<UID>,<PWD>,<DB>);
    if($db->connect_errno > 0){
        die('Unable to connect to database [' . $db->connect_error . ']');
    }

    $sql = "select * from table where ".$col." LIKE '%{$key}%'";

if(!$result = $db->query($sql)){
    die('There was an error running the query [' . $db->error . ']');
}
while($row = $result->fetch_assoc()){
    $array[] = $row['username'];

    $array[] = $row[$term];
}
echo json_encode($array);
?>