Php 通过单击不同的按钮显示不同的表

我有3个不同的表，我想用3个不同的按钮显示，但我不知道是否可以使用php和sql？到目前为止，我一直在挣扎，但我似乎想不出其中的逻辑。 “坦普广场”、“坦普1”和“世纪广场”是我的桌子

我的PHP和sql是最基本的，任何帮助将不胜感激

<form name="type" id="myForm" action="index.php" method="post">
<input type="submit" name="tamp_mall" value="tamp_mall"> ;
<input type="submit" name="tamp1" value="tamp1"> ;
<input type="submit" name="century_square" value="century_square"> ;
</form>

/* Attempt MySQL server connection. Assuming you are running MySQL
server with default setting (user 'root' with no password) */
$link = mysqli_connect("localhost", "root", "", "csv_db");
$tmall = "tamp_mall";
$tamp1 = "tamp1";
$square = "century_square";

// Check connection
if ($link === false) {
die("ERROR: Could not connect. " . mysqli_connect_error());
}

// Attempt select query execution
$sql = "SELECT  KEYWORD, COUNT(*) Count_Duplicate
    FROM $square
    GROUP BY KEYWORD
    HAVING COUNT(*) > 1
    ORDER BY COUNT(*) DESC
    LIMIT 5";

if ($result = mysqli_query($link, $sql)) {
if (mysqli_num_rows($result) > 0) {
    echo "<table>";
    echo "<tr>";
    echo "</tr>";

    while($row = mysqli_fetch_array($result)){
        echo "<tr>";
        echo "<td>" . $row['KEYWORD'] .  "</td>";         
        echo "</tr>";
    }

    echo "</table>";
    // Close result set
    mysqli_free_result($result);
  } else {
    echo "No records matching your query were found.";
  }
} else {
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}

// Close connection
mysqli_close($link);

---

创建onclick事件，在该事件中发送AJAX请求以获取该按钮上所需表的数据，然后以表格式显示结果

<button class = 'btn1'>B1</button>
<button class = 'btn2'>B1</button>
<button class = 'btn3'>B1</button>
<div class='disp_tab'></div>
<script>
   $('.btn1').click(function(){
     ajaxRequest('tab1');
   });
   $('.btn2').click(function(){
     ajaxRequest('tab2');
   });
   $('.btn3').click(function(){
     ajaxRequest('tab3');
   });
   function ajaxRequest(tab){
      $.ajax({
         method : 'POST', //can be GET or post
         url : 'foo/foo/xyz.php', // URL to your php page
         data : {data : tab}, // passing which tab to be displayed
         dataType : 'json', //Data type is json
         success : function(data){
            //fetched data is processed to print in table format inside the div
         },
         error:function(x,y,z){
            //error handling
         },
      });
   }
</script>

PHP:foo/foo/xyz.PHP

<?php 
$tab= $_POST['data'];
switch($tab){
case 'tab1':
//fetch content of table 1 into an array say $arr
echo json.encode($arr);
break;
case 'tab2':
//fetch content of table 2 into an array say $arr
echo json.encode($arr);
break;
case 'tab3':
//fetch content of table 4 into an array say $arr
echo json.encode($arr);
break;
}
?>

---

希望这对你有帮助

请检查下面我的逻辑

PHP:

JS:

---

我不想使用AJAX，因为不需要在点击按钮时启动查询。

非常简单。在php文件中，尝试以下操作：

//write connection code here like
/* Attempt MySQL server connection. Assuming you are running MySQL
server with default setting (user 'root' with no password) */
$link = mysqli_connect("localhost", "root", "", "csv_db");
$tmall = "tamp_mall";
$tamp1 = "tamp1";
$square = "century_square";

// Check connection
if ($link === false) {
die("ERROR: Could not connect. " . mysqli_connect_error());
}

if($_POST["tamp_mall"]){
  //write you code to display from 'tamp_mall' table here
}elseif($_POST["tamp1"]){
  //write you code to display from 'tamp1' table here
}else{
  //write you code to display from 'century_square' table here
}

---

希望这有帮助。

Hi，我应该把js放在哪一部分？我应该把它放在同一个php文件下吗？或者我必须创建一个单独的js文件并将其包含在php文件中吗？不，您可以在同一个php中使用。无需为此创建另一个文件。仅供参考，包括JQ文件。我执行了代码，但我得到的只是来自“$tamp_mallData”表的数据来自“$tamp1Data”表的数据来自“$century_squareData”表的数据来自这里，3按钮您是否按照我的指示只是为了获取信息，包括JQ文件。请检查浏览器控制台是否存在JS错误。也许它说$不是一个函数。这意味着您需要添加JQuery。

<script src='http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js'></script>
<script>

//Initially hide every div except first
$('.btnDiv').hide();
$('#tamp_mall').show();

//Hide/show div 'tamp_mall'
$('#btn_mall').on('click', function (){
    $('.btnDiv').hide();
  $('#tamp_mall').show();
});

//Hide/show div 'tamp1'
$('#btn_tmp1').on('click', function (){
    $('.btnDiv').hide();
  $('#tamp1').show();
});

//Hide/show div 'century_square'
$('#btn_square').on('click', function(){
    $('.btnDiv').hide();
  $('#century_square').show();
});

</script>

//write connection code here like
/* Attempt MySQL server connection. Assuming you are running MySQL
server with default setting (user 'root' with no password) */
$link = mysqli_connect("localhost", "root", "", "csv_db");
$tmall = "tamp_mall";
$tamp1 = "tamp1";
$square = "century_square";

// Check connection
if ($link === false) {
die("ERROR: Could not connect. " . mysqli_connect_error());
}

if($_POST["tamp_mall"]){
  //write you code to display from 'tamp_mall' table here
}elseif($_POST["tamp1"]){
  //write you code to display from 'tamp1' table here
}else{
  //write you code to display from 'century_square' table here
}