我想将从数据库中获取的值数组作为参数传递给php函数。如果我回显该值,它就不会显示它。
<?php
include "config.php";
$sql="select * from project where comp_id='1'";
$sql1=mysql_query($sql);
$rows=array();
while( $fet=mysql_fetch_array($sql1))
{
$rows[]=$fet;
}
echo fun_parameter($rows);
function fun_parameter($rows)
{
echo" rows". $rows['start_date']."values"; // not working
}
foreach($rows as $row)
{
echo $name=$row['start_date']; /working
}
?>
答案 0 :(得分:0)
简单循环可以给出结果,
function fun_parameter($rows)
{
foreach($rows as $row)
{
echo $row['p_id']; /working
echo" rows". $row['start_date']."values";
}
//echo" rows". $rows['start_date']."values"; // not working
}
如果你想从多维数组中获取特定的列,那么看一下php中的array_column()
答案 1 :(得分:-1)
在您的代码中,您尝试在没有指定索引的情况下回显数组$row
,您需要循环遍历该数组,因为它具有多个值,请使用以下代码:
<?php
include "config.php";
$sql="select * from project where comp_id='1'";
$sql1=mysql_query($sql);
$rows=array();
while( $fet=mysql_fetch_array($sql1))
{
$rows[$i]['start_date']=$fet['start_date'];
}
echo fun_parameter($rows);
function fun_parameter($rows)
{
foreach($rows as $row)
{
echo" rows". $row['start_date']."values"; // use loop it will work
}
}
foreach($rows as $row)
{
echo $name=$row['start_date']; // i left this foreach here because you had it in your code
}