<form style="float: right;margin-top: -10px" id="form" onsubmit="return false" action="##" method="post">
<input type="text" style="height: 30px;padding-left: 10px" value="" placeholder="search name" name="name">
<button style="height: 35px;border:0;background-color: #990000;color: #FFFFFF;border-radius: 10px;padding-left: 20px;padding-right: 20px" onclick="search()">search</button>
</form>
</span>
<div id="app" style="position: absolute;top: 50px;background-color: #990000;color:#FFFFFF;width: 70%;text-align: center;z-index: 99999">
<div class="content">
</div>
</div>
ajax提交并且jquery渲染到页面上:
<script>
function search() {
$.ajax({
type: "POST",
dataType: "json",
url: "../includes/dosearch.php",
data: $('#form').serialize(),
success: function (data) {
$.each(data, function(i, n){
$(".content").prepend('<br><a href="demo.php?o_id='+n.o_id+'" style="color: #FFFFFF;"><event></event> name: '+n.name+'</a><br><br>')
})
},
});
}
</script>
PHP返回JSON
$servername = "localhost";
$username = "root";//数据库用户名
$password = "root";//数据库密码
$dbname = "dbname";
$conn = mysqli_connect($servername,$username,$password,$dbname);
if ($conn->connect_error) {
die("Connection failed: " .$conn->connect_error);
}
$search_name = $_POST['name'];
$sql = "SELECT o_id,name FROM organizations WHERE name LIKE '%$search_name%'";
$result = mysqli_query($conn, $sql);
if(!$result) {
echo $sql;
} else{
while($data = mysqli_fetch_assoc($result)) {
$data_search[] = $data;
}
}
//var_dump($data_search);
$data_json = json_encode($data_search);//return to json data
echo $data_json;//output data
效果图: