从 sql 中选择数据以创建缩略图
Selecting data from sql to create a thumbnail
我正在创建一个带有标题、图片和说明的缩略图。我正在尝试 select 来自 table 的数据以将其显示到我的主页中。谁能帮我在我的 php 中创建一个包含我的 sql 详细信息的普通缩略图。我尝试搜索但找不到如何使用 php 而不是 html 创建缩略图。
$sql = "SELECT * FROM news";
if($result = mysqli_query($link, $sql)){
if(mysqli_num_rows($result) > 0){
echo "<table>";
echo "<tr>";
echo "<th>id</th>";
echo "<th>first_name</th>";
echo "<th>last_name</th>";
echo "<th>email</th>";
echo "</tr>";
while($row = mysqli_fetch_array($result)){
echo "<tr>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row['title'] . "</td>";
echo "<td>" . $row['image'] . "</td>";
echo "<td>" . $row['caption'] . "</td>";
echo "</tr>";
}
echo "</table>";
// Free result set
mysqli_free_result($result);
} else{
echo "No records matching your query were found.";
}
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
简而言之,有没有办法从中创建一个 php 文件?
<div class="col-md-4">
<div class="thumbnail">
<img alt="Memory" img src="../../images/a2.jpg">
<div class="caption">
<h3><b>
Title
</b></h3>
<p>
Caption Caption Caption Caption Caption
</p>
<p align="right">
<a class="btn btn-primary" href="news2.html">Read More</a>
</p>
</div>
</div>
</div>
是否要用该模板结构替换 table 结构?您需要调整一些数据来填充超链接(我不知道您要如何构建它)。
while($row=mysqli_fetch_array($result)){
echo "<div class=\"col-md-4\">";
echo "<div class=\"thumbnail\">";
echo "<img alt=\"Memory\" src=\"../../images/{$row["image"]}\">";
echo "<div class=\"caption\">";
echo "<h3>{$row["title"]}</h3>";
echo "<p>{$row["caption"]}</p>";
echo "<p align=\"right\">";
echo "<a class=\"btn btn-primary\" href=\"news2.html\">Read More</a>";
echo "</p>";
echo "</div>";
echo "</div>";
echo "</div>";
}
我正在创建一个带有标题、图片和说明的缩略图。我正在尝试 select 来自 table 的数据以将其显示到我的主页中。谁能帮我在我的 php 中创建一个包含我的 sql 详细信息的普通缩略图。我尝试搜索但找不到如何使用 php 而不是 html 创建缩略图。
$sql = "SELECT * FROM news";
if($result = mysqli_query($link, $sql)){
if(mysqli_num_rows($result) > 0){
echo "<table>";
echo "<tr>";
echo "<th>id</th>";
echo "<th>first_name</th>";
echo "<th>last_name</th>";
echo "<th>email</th>";
echo "</tr>";
while($row = mysqli_fetch_array($result)){
echo "<tr>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row['title'] . "</td>";
echo "<td>" . $row['image'] . "</td>";
echo "<td>" . $row['caption'] . "</td>";
echo "</tr>";
}
echo "</table>";
// Free result set
mysqli_free_result($result);
} else{
echo "No records matching your query were found.";
}
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
简而言之,有没有办法从中创建一个 php 文件?
<div class="col-md-4">
<div class="thumbnail">
<img alt="Memory" img src="../../images/a2.jpg">
<div class="caption">
<h3><b>
Title
</b></h3>
<p>
Caption Caption Caption Caption Caption
</p>
<p align="right">
<a class="btn btn-primary" href="news2.html">Read More</a>
</p>
</div>
</div>
</div>
是否要用该模板结构替换 table 结构?您需要调整一些数据来填充超链接(我不知道您要如何构建它)。
while($row=mysqli_fetch_array($result)){
echo "<div class=\"col-md-4\">";
echo "<div class=\"thumbnail\">";
echo "<img alt=\"Memory\" src=\"../../images/{$row["image"]}\">";
echo "<div class=\"caption\">";
echo "<h3>{$row["title"]}</h3>";
echo "<p>{$row["caption"]}</p>";
echo "<p align=\"right\">";
echo "<a class=\"btn btn-primary\" href=\"news2.html\">Read More</a>";
echo "</p>";
echo "</div>";
echo "</div>";
echo "</div>";
}