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（$field5name）。我尝试在其中添加“onerror”函数但不起作用，我的代码中缺少什么？<!DOCTYPE html><html> <body> <!-- [SEARCH FORM] --> <form method="post" action="1-form.php"> <h1>SEARCH FOR USERS</h1> <input type="text" name="search" required/> <input type="submit" value="Search"/> </form><br/><div align="left"><Font size="4"><a href="logout.php" style="margin-right:80px;">LOGOUT</a></font></div><br> <?phpif (isset($_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>'; } $result->free(); } else { echo '<FONT COLOR="RED" SIZE="5"> NO RESULTS FOUND </FONT>'; }}?> </body></html>
1 回答
弑天下
TA贡献1818条经验 获得超8个赞
您需要做的就是在代码中添加一些测试，看看相关字段是否为空。不确定对你来说什么是空的所以我假设''
if (count($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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