在php [html]中从数据库中显示网格中的数据

时间:2016-04-21 13:03:12

标签: javascript php html css

这是网格的模板,但它不是静态数据库所以我不知道循环是否会起作用,如果它不知道如何使用它

<div class="container">
<div class="row-fluid">
  <ul class="thumbnails">
    <li class="span3">
    <div class="thumbnail" style="padding: 0">
      <div style="padding:4px">
        <img alt="300x200" style="width: 100%" src="http://placehold.it/200x150">
      </div>
      <div class="caption">
        <h2>Fadahunsi Simeon</h2>
        <p>My project description</p>
        <p><i class="icon icon-map-marker"></i> Place, Country</p>
      </div>
      <div class="modal-footer" style="text-align: left">
        <div class="progress" style="background: #ddd">
          <div class="bar bar-danger" style="width: 30%;"></div>
        </div>
        <div class="row-fluid">
          <div class="span4"><b>60%</b><br/><small>FUNDED</small></div>
          <div class="span4"><b>$1000</b><br/><small>PLEDGED</small></div>
          <div class="span4"><b>NOT FUNDED</b><br/><small></small></div>
        </div>
      </div>
    </div>
  </li>
  </ul>
</div>

so lets say that i select all plumbers from the database how will i show each in the grid with their details(link for how the db looks like)

Link for the grid

1 个答案:

答案 0 :(得分:0)

您可以在php中使用循环来创建这些容器,以便显示详细信息。

示例代码:

$row = mysqli_fetch_array($result)
while($row){
    echo $row['name']. " - ". $row['age'];
    echo "<br />";
}

以这种方式循环并创建这些容器。

样品MYSQL&amp; PHP连接代码:

<?php
if(isset($_POST['formSubmit'])){
        $DB_USER="root";
        $DB_PASS="";
        $DB_HOST="localhost";
        $DB_NAME="contacts";

        $conn = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME);
        if ($conn->connect_error) {
            die("Connection failed: " . $conn->connect_error);
        } 

        $name = $_POST['formName'];
        $title = $_POST['formTitle'];
        $enquiry = $_POST['formEnquiry'];

        $sql = "INSERT INTO enquiries (Name, Title, enquiry) VALUES('".$name."', '".$title."', '".$enquiry."')";

        $result = $conn->query($sql);
        if($result){
            alert("Thank You");
        }else{
            alert("error!");
        }
        mysqli_close($conn);


}
?>