使用 HTML 复选框在 PHP 代码中动态创建 SQL 查询

Question

我目前有一个 html / PHP 网页，它使用一系列用户输入来查询 SQL 数据库并返回数据以进行绘图。用户输入由 html 复选框驱动，由 SQL 数据库中的数据驱动并使用 PHP 代码创建。整个 html 代码太大，这里放不下，但下面是我想问的部分的摘录......

...

<div id="AccProject" class="w3-hide">

<?php 
  $queryProject = "SELECT DISTINCT ProjID, Project, Sample, Type, Description FROM PROJECTS ORDER BY Project ASC";
  $resultProject = $dbhandle->query($queryProject) or exit("Error code ({$dbhandle->errno}): {$dbhandle->error}");
  while($row=mysqli_fetch_array($resultProject)){
    echo "<input id='".$row['ProjID']."' value='".$row['ProjID']."' name='projID[]' type='checkbox' class='chkbox'>";
    echo "<label title='".$row['Description']."&#13;&#13;Project: ".$row['Project']."&#13;Sample: ".$row['Sample']."&#13;Type: ".$row['Type']."' for='".$row['ProjID']."' class='chkbox-label'> ".$row['Project']." | ".$row['Sample']." | ".$row['Type']."</label>";
  }
?>

</div>

...

可以看到我查询了SQL数据库，返回了一些字段，然后用PHP生成了一些复选框。后来在html中，我使用javascript提取这些复选框的选中/未选中状态并再次查询数据库以返回更多数据。

我想做的是添加一些预过滤器，这将减少页面上用户可见的复选框数量。例如，如果“类型”字段定义数据条目是“蛋糕”、“饼干”还是“布丁”，我想添加一些开关，然后修改 SQL 查询。我在下面给出了一个带有一些伪代码的示例。

...

<div id="AccProject" class="w3-hide">

<label class="form-switch"><input type="checkbox" id="filterCake"><i></i> Select Cakes only </label>
<label class="form-switch"><input type="checkbox" id="filterBiscuit"><i></i> Select Biscuits only </label>
<label class="form-switch"><input type="checkbox" id="filterPudding"><i></i> Select Puddings only </label>

<?php 

  if filterCake == True
    $queryProject = "SELECT DISTINCT ProjID, Project, Sample, Type, Description FROM PROJECTS WHERE Type == 'Cake' ORDER BY Project ASC";

  if filterBiscuit == True
    $queryProject = "SELECT DISTINCT ProjID, Project, Sample, Type, Description FROM PROJECTS WHERE Type == 'Biscuit'  ORDER BY Project ASC";

  if filterPudding == True
    $queryProject = "SELECT DISTINCT ProjID, Project, Sample, Type, Description FROM PROJECTS WHERE Type == 'Pudding' ORDER BY Project ASC";


  $resultProject = $dbhandle->query($queryProject) or exit("Error code ({$dbhandle->errno}): {$dbhandle->error}");
  while($row=mysqli_fetch_array($resultProject)){
    echo "<input id='".$row['ProjID']."' value='".$row['ProjID']."' name='projID[]' type='checkbox' class='chkbox'>";
    echo "<label title='".$row['Description']."&#13;&#13;Project: ".$row['Project']."&#13;Sample: ".$row['Sample']."&#13;Type: ".$row['Type']."' for='".$row['ProjID']."' class='chkbox-label'> ".$row['Project']." | ".$row['Sample']." | ".$row['Type']."</label>";
  }
?>

</div>

...

如何使用我的代码实现这一点？

我需要做的就是根据蛋糕、饼干或布丁复选框的状态更改 PHP 代码中的 SQL 查询（添加 WHERE 子句），但我一直在努力寻找如何做到这一点。

非常感谢任何帮助！

Answer 1

您必须使用 AJAX 执行此操作，并且每次用户单击复选框时，都会将值发送到不同的页面，然后您可以在另一个页面中使用 PHP 循环遍历结果并以 HTML 形式返回或返回结果并在复选框页面上使用JS循环遍历它

看一个例子：

 $("#formName").on("change", "input:checkbox", function(){
   //send $(this).val() with ajax
   //Assuming you are selecting cakes and
   //getting  back an array with cakes
   //you are the looping thru the array and gettig populating the html
   
   
   var response = [];
   var cakes = {};
   cakes.description = 'Cake Description 1';
   cakes.project = 'Cake Project 1';
   response.push(cakes);
   cakes = {};
   cakes.description = 'Cake Description 2';
   cakes.project = 'Cake Project 2';
   response.push(cakes);
   
   
   var formattedResponse = '';
   response.forEach(function(res){
     formattedResponse += `This is  ${res.description} <br>`});
   
   
    $('#responseContainer').html(formattedResponse);
  
    });

<script src="https://code.jquery.com/jquery-2.2.4.js"></script>

<form action="" id="formName">
  

<label class="form-switch"><input type="checkbox" value="cake" id="filterCake"><i></i> Select Cakes only </label>
<label class="form-switch"><input type="checkbox" value="biscuits" id="filterBiscuit"><i></i> Select Biscuits only </label>
<label class="form-switch"><input type="checkbox" value="puddings" id="filterPudding"><i></i> Select Puddings only </label>

  </from>


<div id="responseContainer"></div>

Answer 2

是的，您可以通过动态查询从数据库中获取数据。请试试这个。

$type = $post['type']; //'蛋糕','饼干','布丁'

$queryProject = "SELECT DISTINCT ProjID, Project, Sample, Type, Description FROM PROJECTS WHERE Type = '" 。 $type ."' ORDER BY Project ASC";