注意:数组到字符串的转换-PHP& MySQL的 [英] Notice: Array to string conversion - PHP & mySQL

问题描述

我一直在阅读与此相关的每个线程，但是我总是弄错了.

I've been reading in every thread in here that is related to this but I always get it wrong.

请帮助，因为我总是收到错误消息

Please help cause I always get the error

"$ address []"行中的注意:数组到字符串的转换" = mysql_result($ row，0);"

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

如下.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);

推荐答案

更改此行

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

对此:

 $address[] = $row;

然后要查看新的$address数组中可用的键和值，可以执行以下操作:

And then to see the keys and values available in the new $address array, you can do something like this:

 print_r($address);

为了保持implode()的功能，请执行以下操作:

In order to keep implode() functional, do something like this:

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";
}
}

希望有帮助...

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