Question

我有一个搜索框，我将在其中搜索名称，如果该名称存在于数据库中，则应获取该名称并将其显示在选择框中。我无法做到这一点。

<input type="text" class="form-control searchbox" name="search" id="searchbox">
<div id="re"></div>

$(document).on("keyup", ".searchbyname", function() {
  var searchname = $(this).val();
  console.log(searchname);

  if (searchname != '') {
    $.ajax({
      type: "POST",
      url: "search.php",
      data: {
        searchname: searchname
      },
      success: function(html) {
        $('#re').html(html);
      }
    });
  }
  return false; 
});

$name = $_POST['searchname'];
$sql_res = $conn->query("select *from tbl where firstName like '".$name."%'");
if ($sql_res->num_rows > 0)
{
  while ($row = $sql_res->fetch_assoc()) {
    $row['name'];//here how to send back name in select box i am not able to do so  
  }
}

Answer 1

您需要使用echo将数据发送回php。在您的php代码中，如下所示：

  if($sql_res->num_rows > 0)
      {
     echo " <select name='na'>";
     echo "<option value=''>Select</option>";
     //if names exists print the value in option
         while($row=$sql_res->fetch_assoc())   
                {
                echo "<option value='". $row['id']."'>".$row['name']."</option>"; //here printing name
                }
                        echo "</select>";


    }

如何使用PHP和AJAX显示数据库中的数据匹配？

1 个答案: