How to implement the like function in php mysql-PHP Problem-php.cn

How to implement the like function in php mysql

angryTom

Release： 2023-02-27 07:42:01

Original

4357 people have browsed it

php How to implement the like function in mysql

1. index.php reads database likes Number of tables.

<!DOCTYPE html>
<html>
<head>
	<meta charset="utf-8">
	<title>点赞</title>
	<script src="jquery-2.1.1.min.js"></script>
	<script src="index.js"></script>
</head>
<body>
	<button id="btn">赞</button>
	<span id="result">
		<?php
		$con = mysqli_connect(&#39;localhost&#39;, &#39;username&#39;, &#39;password&#39;);
		if(! $con )
		{
		    die(&#39;连接失败: &#39; . mysqli_error($con));
		}
		mysqli_select_db($con,&#39;jwhuang&#39;);
		mysqli_query($con,"set names utf8");
		$result = mysqli_query($con,"SELECT * FROM zan");
		while($row = mysqli_fetch_array($result))
		{
		echo $row[&#39;zan&#39;];
		}
		mysqli_close($con)
		?>
	</span>
</body>
</html>

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2. server.php When the user clicks the like button, a piece of data will be inserted into the database.

<?php
header("Content-type:text/html;charset=utf-8");
$con = mysqli_connect(&#39;localhost&#39;, &#39;username&#39;, &#39;password&#39;);
if(!$con )
{
    die(&#39;连接失败: &#39; . mysqli_error($con));
}
mysqli_select_db($con,&#39;jwhuang&#39;);
mysqli_query($con,"set names utf8");
mysqli_query($con,"UPDATE zan SET zan = zan+1");
$result = mysqli_query($con,"SELECT * FROM zan");
if(isset($_GET[&#39;name&#39;])){
while($row = mysqli_fetch_array($result))
  {
  echo $row[&#39;zan&#39;];
  }
}else{
	echo "赞失败！";
}
mysqli_close($con);
?>

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3. index.js request data

$(document).ready(function(){
	$("#btn").on("click",function(){
		$.get("sever.php",{name:$("#btn").val()},function(data){
			$("#result").text(data);
		});
	});
});

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