PHP的Ajax創建像谷歌搜索框

Question

我想創建一個搜索框，類似於Google搜索框。

我的數據庫中有一個表，其中存儲搜索文本。

---------------------------
| searchdata     | output |
---------------------------
| id             | id=-1  |
| Data for id 1  | id=1   |
| data for 2     | id=2   |
| datatx id 2    | id=4   |
| datacv for id  | id=5   |
---------------------------

現在在我的php文件中

<?php

        if ($con = mysqli_connect("localhost","root","root","search")) {


          $chatData1 = $_POST['request'];


$textm= str_replace(" ", "%' or searchdata like '%", $chatData1);


          if (mysqli_connect_errno())
           {
               echo "Failed to connect to MySQL: " . mysqli_connect_error();
             }

$result = mysqli_query($con,"SELECT * FROM  search  where searchdata like '%{$textm}%' ");

if($row = mysqli_fetch_array($result))
  {
 echo $row['output'];


}
 else{
echo "Please use English(US) Language.";

}


}
mysqli_close($con);

?>

例如，如果用戶data for id輸入data for id則sql查詢必須像這樣

SELECT * 
FROM search  
WHERE searchdata LIKE '%data%' OR 
      searchdata LIKE '%for%' OR 
      searchdata LIKE '%id%'`

因此它必須給出輸出acc。 到表id=1 ，但它給我輸出id=-1

Answer 1

您可以使用jQuery Ui Autocomplete解決此問題，只需指定字段並使用JSON顯示數據

$data = array()
if($row = mysqli_fetch_array($result))
{
  data[] = $row['output'];
}
// Outputs JSON
print json_encode($data);