实际上我猜这个问题是重复的，但我不明白我的 SQL 代码有什么问题。我的错误是
Notice: Undefined index: CI in C:\wamp\www\LOCATIONVIEWER\exampleDB.php on line 30.
我需要将列的所有十进制值转换为十六进制值。帮我做一下...:)
这是我的代码:
<html>
<head>
<title>test</title>
</head>
<body>
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "locationviewer";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
//CI is a column name of the table
$sql = "SELECT CONV(CI,10,16) FROM locationdata";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "CI: " . $row["CI"]."<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
</body>
</html>
最佳答案
您应该更改以下代码:
$sql = "SELECT CONV(CI,10,16) FROM locationdata";
至
$sql = "SELECT CONV(CI,10,16) AS `CI` FROM locationdata";
发生这个问题是因为 php 无法看到 CI 索引，并且当前索引类似于 CONV(CI,10,16)
关于php - 将表列中的 DEC 值转换为十六进制，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/37268913/