是否有一种方法可以在MySQL中使用PHP获取表的列名?
当前回答
在mysql中,通过关键字或查询来获取列的详细信息和表的结构
1.DESC table_name
2.描述table_name
3.显示table_name中的列
4.创建表table_name
5.解释table_name
其他回答
这个怎么样:
SELECT @cCommand := GROUP_CONCAT( COLUMN_NAME ORDER BY column_name SEPARATOR ',\n')
FROM INFORMATION_SCHEMA.COLUMNS
WHERE TABLE_SCHEMA = 'my_database' AND TABLE_NAME = 'my_table';
SET @cCommand = CONCAT( 'SELECT ', @cCommand, ' from my_database.my_table;');
PREPARE xCommand from @cCommand;
EXECUTE xCommand;
Mysqli fetch_field()为我工作:
if ($result = $mysqli -> query($sql)) {
// Get field information for all fields
while ($fieldinfo = $result -> fetch_field()) {
printf("Name: %s\n", $fieldinfo -> name);
printf("Table: %s\n", $fieldinfo -> table);
printf("Max. Len: %d\n", $fieldinfo -> max_length);
}
$result -> free_result();
}
来源:https://www.w3schools.com/pHP/func_mysqli_fetch_field.asp
在mysql中,通过关键字或查询来获取列的详细信息和表的结构
1.DESC table_name
2.描述table_name
3.显示table_name中的列
4.创建表table_name
5.解释table_name
我写了一个简单的php脚本,通过php获取表列: Show_table_columns.php
<?php
$db = 'Database'; //Database name
$host = 'Database_host'; //Hostname or Server ip
$user = 'USER'; //Database user
$pass = 'Password'; //Database user password
$con = mysql_connect($host, $user, $pass);
if ($con) {
$link = mysql_select_db($db) or die("no database") . mysql_error();
$count = 0;
if ($link) {
$sql = "
SELECT column_name
FROM information_schema.columns
WHERE table_schema = '$db'
AND table_name = 'table_name'"; // Change the table_name your own table name
$result = mysql_query($sql, $con);
if (mysql_query($sql, $con)) {
echo $sql . "<br> <br>";
while ($row = mysql_fetch_row($result)) {
echo "COLUMN " . ++$count . ": {$row[0]}<br>";
$table_name = $row[0];
}
echo "<br>Total No. of COLUMNS: " . $count;
} else {
echo "Error in query.";
}
} else {
echo "Database not found.";
}
} else {
echo "Connection Failed.";
}
?>
享受吧!
我需要列名作为平面数组,而其他答案返回关联数组,所以我使用:
$con = mysqli_connect('localhost',$db_user,$db_pw,$db_name);
$table = 'people';
/**
* Get the column names for a mysql table
**/
function get_column_names($con, $table) {
$sql = 'DESCRIBE '.$table;
$result = mysqli_query($con, $sql);
$rows = array();
while($row = mysqli_fetch_assoc($result)) {
$rows[] = $row['Field'];
}
return $rows;
}
$col_names = function get_column_names($con, $table);
$col_names现在等于:
(
[0] => name
[1] => parent
[2] => number
[3] => chart_id
[4] => type
[5] => id
)
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