如果mysql中没有记录，php显示默认图像
我得到了下面的 php 代码来搜索和显示来自 mysql 的图像，到目前为止工作正常，我想添加一个功能来显示默认图像，如果搜索结果在 mysql 中没有找到记录（$field5name）。
我尝试在其中添加“onerror”函数但不起作用，我的代码中缺少什么？
<!DOCTYPE html>
<html>
<body>
<!-- [SEARCH FORM] -->
<form method="post" action="1-form.php">
<h1>SEARCH FOR USERS</h1>
<input type="text" name="search" required/>
<input type="submit" value="Search"/>
</form>
<br/>
<div align="left"><Font size="4"><a href="logout.php" style="margin-right:80px;">LOGOUT</a></font></div><br>
<?php
if (isset($_POST['search'])) {
require "2-search.php";
echo '<table border="1" cellspacing="2" cellpadding="2">
<tr>
<td> <font face="Arial">ID</font> </td>
<td> <font face="Arial">OTP</font> </td>
<td> <font face="Arial">EMAIL</font> </td>
<td> <font face="Arial">STATUS</font> </td>
<td> <font face="Arial">TIMESTAMP</font> </td>
</tr>';
// DISPLAY RESULTS
if (count($results) > 0) {
foreach ($results as $r) {
$field1name = $r["id"];
$field2name = $r["otp"];
$field3name = $r["email"];
$field4name = $r["is_expired"];
// $field5name = $r["create_at"];
$field5name = $r["jpeg_info"];
echo '<tr>
<td>' . $field1name . '</td>
<td>' . $field2name . '</td>
<td>' . $field3name . '</td>
<td>' . $field4name . '</td>
<td><a href = ' . $field5name . ' target = "_blank">
<img src=' . $field5name . ' height="100" width="100">
</td> //<------------------- i am stuck here
</tr>';
}
$result->free();
} else {
echo '<FONT COLOR="RED" SIZE="5"> NO RESULTS FOUND </FONT>';
}
}
?>
</body>
</html>
子衿沉夜1回答
-
弑天下
您需要做的就是在代码中添加一些测试，看看相关字段是否为空。不确定对你来说什么是空的所以我假设''if (count($results) > 0) { foreach ($results as $r) { $field1name = $r["id"]; $field2name = $r["otp"]; $field3name = $r["email"]; $field4name = $r["is_expired"]; // $field5name = $r["create_at"]; $field5name = $r["jpeg_info"]; if ( $field5name == '' ) { $img = 'default.png'; } else { $img = $field5name; } echo "<tr> <td>$field1name</td> <td>$field2name</td> <td>$field3name</td> <td>$field4name</td>"; <td> <a href='$img' target='_blank'> <img src='$img' height='100' width='100'> </a> </td> </tr>"; }}
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