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

这对我很有效。

$sql = "desc MyTableName";
$result = @mysql_query($sql);
while($row = @mysql_fetch_array($result)){
    echo $row[0]."<br>";
}

其他回答

你可能还想检查mysql_fetch_array(),如下所示:

$rs = mysql_query($sql);
while ($row = mysql_fetch_array($rs)) {
//$row[0] = 'First Field';
//$row['first_field'] = 'First Field';
}

有趣的是,您可以使用 EXPLAIN table_name与DESCRIBE table_name和SHOW COLUMNS FROM table_name同义 尽管EXPLAIN更常用来获取关于查询执行计划的信息。

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

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;

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!