Question

我想创建一个搜索框，例如Google搜索框。

我的数据库中有一个表格，其中存储了搜索文本。

---------------------------
| 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   |
---------------------------

现在在我的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，如果用户键入data for id sql查询必须像这样

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

所以它必须给出输出acc。表id=1，但它给我输出id=-1

Answer 1

您可以使用jQuery Ui Autocomplete解决此问题，只需指定字段并使用JSON显示数据

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

php ajax创建一个像谷歌一样的搜索框

1 个答案: