php - 如何在实时搜索栏中创建可点击的搜索结果

Question

我的实时搜索栏以下列方式显示搜索结果：

[{"item_name":"johnnie-walker-odyssey","href":"profile.php?item_name=johnnie-walker-odyssey"},{"item_name":"jack daniels tennessee-fire","href": “profile.php?item_name=jack daniels tennessee-fire”}。

我打算将结果显示为链接，这样当用户点击他想要的结果时，他们可以被重定向到另一个显示结果详细信息的页面。

<?php   
$link = mysqli_connect("localhost", "root", "charles", "shoppingcartdemo");

// Check connection
if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}

// Escape user inputs for security
$term = mysqli_real_escape_string($link, $_REQUEST['term']);

if(isset($term)){
// Attempt select query execution
$sql = "SELECT * FROM shopping_items WHERE item_name LIKE '" . $term . "%'";
if($result = mysqli_query($link, $sql)){
if(mysqli_num_rows($result) > 0){
while($row = mysqli_fetch_array($result)){

//display search results links   
$array [] = array("item_name"=> $row['item_name'],"href"=>"profile.php? 
item_name=".$row['item_name']);}
echo json_encode($array)
// Close result set
mysqli_free_result($result);
} else{
echo "<p>No matches found</p>";
}
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}


}


// close connection
mysqli_close($link);
?>

标签： phpjqueryajaxtypeahead.js