注意:数组到字符串的转换-PHP& MySQL的 [英] Notice: Array to string conversion - PHP & mySQL
本文介绍了注意:数组到字符串的转换-PHP& MySQL的的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
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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