php - 通知 : Array to string conversion - PHP & mySQL

Question

我一直在阅读这里与此相关的每个线程，但我总是弄错。

请帮忙，因为我总是得到错误

"Notice: Array to string conversion" in line "$address[] = mysql_result($row, 0 );"

下面。请帮忙。

if ($p_address=mysql_query($email))
{
$address = array();

while($row = mysql_fetch_assoc($p_address))
{     
 $address[] = mysql_result($row, 0 );
}  

$all_address = implode(',', $address);

Answer 1

改变这一行

 $address[] = mysql_result($row, 0 );

对此:

 $address[] = $row;

然后要查看新的 $address 数组中可用的键和值，您可以这样做:

 print_r($address);

为了保持 implode() 的功能，做这样的事情:

for ($i = 0; $i < count($address); $i++) {
  $all_address[] = implode(',', $address[$i]);
}

最终输出:

if ($p_address=mysql_query($email))
{
$address = array();

while($row = mysql_fetch_assoc($p_address))
{     
 $address[] = $row;
}

for ($i = 0; $i < count($address); $i++) {
  $all_address[] = implode(',', $address[$i]);
}

// Example for outputting on screen:
foreach ($all_address as $aa) {
  print $aa . "<br/>\n";
}
}

希望对您有所帮助...