Question

我使用phpLiteAdmin创建了一个数据库，它有很多条目，但它们都是文本或数字。

我想知道我是否可以在此数据库中添加图像部分，或者我该怎么做。我认为我可以做到的另一种方式是使用href标签并将其移动到位。必须有更好的方法来做到这一点。

这是我的php和html的代码。

<head>
   <link type="text/css" rel="stylesheet" href="Car_Style.css"/>
</head>
<body>
    <header id="header" class="loading" style="opacity: 1";]>
    <!-- Logo -->
      <div id="logo"></div>
      <!-- Nav -->
        <nav id="nav">
            <ul>
                <li><a href="#intro">header</a></li>
                <li><a href="#one">Header</a></li>
                <li><a href="#two">header</a></li>
                <li><a href="#work">My Header</a></li>
                <li><a href="#contact">Header</a></li>
            </ul>
        </nav>
    </header>
    <!-- This is where all the images go from each of the entries -->
</body>

<?php

    try {
      # Connect to SQLite database
      $dbh = new PDO("sqlite:Car_Sales_Network");

      # Prepare SQL statement
      $sth = $dbh->prepare('SELECT * FROM Cars_On_Network' );

     # Set the Fetch Mode to Associative Array
     $sth->setFetchMode(PDO::FETCH_ASSOC);

     # Run the query on the database
     $sth->execute();

    //table printout
    echo "<table>";

    # Loop through returned records
    while($row = $sth->fetch()) {
      //print_r($row);spot
      //print_r($row);

    echo "<table>";
    echo "<th>Car Make:</th>";
    echo "<th>Car Model:</th>";
    echo "<th>Car Badge:</th>";
    echo "<th>Car Price:</th>";
    echo "<th>Car Transmission:</th>";
    echo "<th>Car P Plate Legality:</th>";
    echo "<div id='1'>";
    echo "<tr>";
    echo "<td>".  $row["car_make"] . "</td>";
    echo "<td>".  $row["car_model"] . "</td>";
    echo "<td>".  $row["car_badge"] . "</td>";
    echo "<td>".  $row["price"] . "</td>";
    echo "<td>".  $row["trans"] . "</td>";
    echo "<td>".  $row["P_Plate_Legal"] . "</td>";
    echo "</tr>";
    echo "</div>";

    //echo $row["Game_ID"];
    echo "<br>";
  }

  echo "</table>";

 } 
 catch(PDOException $e) {
    echo $e->getMessage();
}

?>

这是我的CSS。

* {
    font-family: Arial;
    color: black;
 }

table {
    padding: 20px;
}

th td {
   padding-right:  20px;
   padding-left: 20px;
   padding-top: 20px;
   padding-bottom: 20px;
}


/*********************************************************************************/
/* Header                                                                        */
/*********************************************************************************/
#header ul li {
    color:white;
    text-decoration: underline;
}

#header {
    position: fixed;
    z-index: 10000;
    left: 0;
    top: 0;
    width: 100%;
    background: #000000;
    background: rgba(0, 0, 0, 0.95);
    height: 3em;
    line-height: 3em;
    box-shadow: 0 0 0.15em 0 rgba(0,0,0,0.1);
}

body {
    padding-top: 3em;
}

#logo {
    position: absolute;
    left: 1em;
    top: 0;
    height: 3em;
    line-height: 3em;
    letter-spacing: -1px;
}

#logo a {
    font-size: 1.25em;
}

#nav {
    position: absolute;
    right: 0.5em;
    top: 0;
    height: 3em;
    line-height: 3em;
}

#nav ul {
    margin: 0;
}

#nav ul li {
    display: inline-block;
    margin-left: 0.5em;
    font-size: 0.9em;
}

#nav ul li a {
    display: block;
    color: inherit;
    text-decoration: none;
    height: 3em;
    line-height: 3em;
    padding: 0 0.5em 0 0.5em;
    outline: 0;
}

显然它没有完成，但我怎么能让每辆车都有一张照片。

干杯。

Answer 1

首先，您必须找到存储图像的方法。

我们通常不会将图像的二进制数据存储在数据库中。相反，我们将图像文件存储在服务器（或第三方CDN）的某个位置。
然后我们在数据库中创建一个新的varchar(255)并存储这些图像的网址（例如http://foobar.com/some/image.jpg）。如需解释，请说我们会在上表中添加一列car_image_url。

让我们稍微更改PHP代码以添加显示列：

<link type="text/css" rel="stylesheet" href="Car_Style.css"/>

</head>

<body>
    <header id="header" class="loading" style="opacity: 1";]>

                <!-- Logo -->
                    <div id="logo">

                    </div>

                <!-- Nav -->
                    <nav id="nav">
                        <ul>
                            <li><a href="#intro">header</a></li>
                            <li><a href="#one">Header</a></li>
                            <li><a href="#two">header</a></li>
                            <li><a href="#work">My Header</a></li>
                            <li><a href="#contact">Header</a></li>
                        </ul>
                    </nav>

            </header>

    <!-- This is where all the images go from each of the entries -->



</body>




<?php

try {
  # Connect to SQLite database
  $dbh = new PDO("sqlite:Car_Sales_Network");

  # Prepare SQL statement
  $sth = $dbh->prepare('SELECT * FROM Cars_On_Network' );

  # Set the Fetch Mode to Associative Array
  $sth->setFetchMode(PDO::FETCH_ASSOC);

  # Run the query on the database
  $sth->execute();

 //table printout
  echo "<table>";

  # Loop through returned records
  while($row = $sth->fetch()) {
    //print_r($row);spot
        //print_r($row);

    /* added these line */
    # If the image field is empty, change to dummy image
    if (empty($row['car_image_url'])) {
        $row['car_image_url'] = 'http://yourdomain.com/some/dummy.jpg';
    }
    /* // added these line */


 echo "<table>";
    echo "<th>Car Make:</th>";
    /* added this line */
    echo "<th>Car Image:</th>";
    /* // added this line */
    echo "<th>Car Model:</th>";
    echo "<th>Car Badge:</th>";
    echo "<th>Car Price:</th>";
    echo "<th>Car Transmission:</th>";
    echo "<th>Car P Plate Legality:</th>";


    echo "<div id='1'>";
    echo "<tr>";
    echo "<td>".  $row["car_make"] . "</td>";
    /* added this line */
    echo "<td><img src=\"".  $row["car_image_url"] . "\" /></td>";
    /* // added this line */
    echo "<td>".  $row["car_model"] . "</td>";
    echo "<td>".  $row["car_badge"] . "</td>";
    echo "<td>".  $row["price"] . "</td>";
    echo "<td>".  $row["trans"] . "</td>";
    echo "<td>".  $row["P_Plate_Legal"] . "</td>";


    echo "</tr>";
    echo "</div>";



      //echo $row["Game_ID"];
    echo "<br>";
  }

echo "</table>";


  } 
catch(PDOException $e) {
    echo $e->getMessage();

}

?>

随心所欲地设置CSS样式。

如何使用表格数据添加图像

1 个答案: