我有以下 PHP 5.4 脚本,它从 MySQL 5.7 中选择,并且工作正常。
<?php
$link = new mysqli("localhost","my_username","my_password", "my_schema");
if (mysqli_connect_errno())
{
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$sugg_query =
"SELECT
s.prim_key,
s.created_date,
s.created_by,
s.suggestion
FROM suggestion s;";
if ($sugg_result = mysqli_query($link, $sugg_query))
{
// determine number of rows result set */
$sugg_row_cnt = mysqli_num_rows($sugg_result);
printf("Result set has %d rows.\n", $sugg_row_cnt);
while($sugg_row = mysqli_fetch_array($sugg_result))
{
echo "<br />";
echo
$sugg_row['prim_key'] . " " .
$sugg_row['created_date'] . " " .
$sugg_row['created_by'] . " " .
$sugg_row['suggestion'] . " " ;
}
/* close result set */
mysqli_free_result($sugg_result);
}
//Close the connection
mysqli_close($link);
//close PHP
?>
我想向结果添加列标题。如果我将 SELECT 语句更改为以下内容,我将得到仅生成行计数行的结果。
$sugg_query =
"SELECT
s.prim_key as 'Key',
s.created_date as 'Date',
s.created_by as 'Source',
s.suggestion as 'Suggestion'
FROM suggestion s;";
如何从 PHP 中的 SELECT 获取数据值和列标题?
最佳答案
要获取标题,您可以迭代一行,例如:
foreach($sugg_row as $header => $value){
echo $header;
}
所以也许这样的东西应该在您的代码中工作:
$rows = mysqli_fetch_array($sugg_result)
foreach($rows[0] as $header => $value){
echo $header;
}
您还需要记住,如果没有行,mysqli_fetch_array() 将返回 NULL
关于PHP 从 mysql 中选择列标题名称,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/39856396/