注意：数组到字符串的转换 - PHP & mySQL
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Notice: Array to string conversion - PHP & mySQL
提问by banana
I've been reading in every thread in here that is related to this but I always get it wrong.
我一直在这里阅读与此相关的每个线程，但我总是弄错。
Please help cause I always get the error
请帮助因为我总是收到错误
"Notice: Array to string conversion" in line "$address[] = mysql_result($row, 0 );"
“注意：数组到字符串的转换”行“$address[] = mysql_result($row, 0);”
below. Please help.
以下。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);
采纳答案by jerdiggity
Change this line
改变这一行
$address[] = mysql_result($row, 0 );
To this:
对此：
$address[] = $row;
And then to see the keys and values available in the new $addressarray, you can do something like this:
然后要查看新$address数组中可用的键和值，您可以执行以下操作：
print_r($address);
In order to keep implode()functional, do something like this:
为了保持implode()功能，请执行以下操作：
for ($i = 0; $i < count($address); $i++) {
$all_address[] = implode(',', $address[$i]);
}
Final output:
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";
}
}
Hope that helps...
希望有帮助...
回答by Aris
$row is set in every iteration of the while loop. every time it contains a new table record. So you just need to add each record in the address array.
$row 在 while 循环的每次迭代中设置。每次它包含一个新的表记录。所以你只需要在地址数组中添加每条记录。
while($row = mysql_fetch_assoc($p_address))
{
$address[] = $row;
}
