我正在使用PHP，SQL和JqueryUI进行搜索

Question

我正在使用PHP，SQL和JqueryUI进行搜索，这是网址：http：//digital-axioma.com.co/top100/

目前，用户可以查找名为“Nombre”的输入，但我需要用户可以查找名为“Nombre”的输入或称为“RazónSocial”的输入或称为“Ciudad”的输入。

这是代码：

<?php
	class Conexion extends mysqli {
		public function __construct(){
			parent::__construct("localhost","root","","top");
			$this->query("SET NAMES 'utf8';");
			$this->connect_errno ? die("Error con la conexion") : $x = "Conectado";
			unset($x);
		}

/*The mysqli_fetch_array() function fetches a result row as an associative array, a numeric array, or both.*/
		public function recorrer($y){
			return mysqli_fetch_array($y);
		}

/* The mysqli_num_rows() function returns the number of rows in a result set */
		public function rows($y){
			return mysqli_num_rows($y);
		}
	}
?>

<?php
	include("class.conection.php");

	class Ajax {

		public $buscador;
		public function Buscar($a){
			
			$db = new Conexion();
			$this->buscador = $db->real_escape_string($a);
			$sql = $db->query("SELECT * FROM top100seis WHERE Nombre LIKE '%$this->buscador%';");

			while($array = $db->recorrer($sql)){
				$resultado[] = $array["Nombre"];
			}
			
			return $resultado;
		}
	}

$busqueda = new Ajax();
echo json_encode($busqueda->Buscar($_GET["term"]));
?>

<aside id="buscador">
        <div class="container">
            <form method="GET" id="formulario-busqueda">
                <div class="caja8">
                   <input type="text" name="buscar" id="buscar" placeholder="Buscar" /> 
                </div>
                <div class="caja2">
                    <button type="submit" id="boton-buscar"><i class="fa fa-search" aria-hidden="true"></i></button>
                </div>
            </form>
        </div>
    </aside>

Answer 1

在你的mysql中使用UNION

SELECT * FROM top100seis WHERE Nombre LIKE '%$this->buscador%'
UNION
SELECT * FROM top100seis WHERE Razón Social LIKE '%$this->buscador%'
UNION
SELECT * FROM top100seis WHERE Ciudad LIKE '%$this->buscador%'