Question

我有3个不同的表，我想用3个不同的按钮显示，但我不知道它是否可以使用php和sql？到目前为止，我一直在努力，但我似乎无法想到逻辑。＆＃39; tamp_mall＆＃39;，＆＃39; tamp1＆＃39;，＆＃39; century_square＆＃39;是我的桌子。

我的PHP和sql是最基本的，任何帮助都将不胜感激。

<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);

Answer 1

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

<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;
}
?>

希望这能帮到你！

Answer 2

请在下面查看我的逻辑。

<强> PHP：

<?php
/* 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");


$tamp_mallData = getData('tamp_mall', $link);
$tamp1Data = getData('tamp1', $link);
$century_squareData = getData('century_square', $link);

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


//This function return data from db...
function getData($tblName, $link){
    $returnData = array();
    // Attempt select query execution
    $sql = "SELECT  KEYWORD, COUNT(*) Count_Duplicate
        FROM $tblName
        GROUP BY KEYWORD
        HAVING COUNT(*) > 1
        ORDER BY COUNT(*) DESC
        LIMIT 5";

    if ($result = mysqli_query($link, $sql)) {
        if(mysqli_num_rows($result) > 0) {
        while($row = mysqli_fetch_array($result)){
           $returnData = $row;
        }

        // 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);
    }

    return $returnData;
}

// Close connection
mysqli_close($link);

?>
<div class="btnDiv" id="tamp_mall">Data from '$tamp_mallData' table comes here</div>
<div class="btnDiv" id="tamp1">Data from '$tamp1Data' table comes here</div>
<div class="btnDiv" id="century_square">Data from '$century_squareData' table comes here</div>

<input type="button"  name="tamp_mall" value="tamp_mall" id="btn_mall" />
<input type="button" name="tamp1" value="tamp1" id="btn_tmp1" />
<input type="button" name="century_square" value="century_square" id="btn_square" />

<强> JS：

<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>

我不想使用AJAX，因为没有必要在按钮点击时触发查询。

Answer 3

非常简单。在你的php文件中，尝试类似的东西;

//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
}

希望这会有所帮助..

单击不同的按钮显示不同的表

3 个答案: