Php 在以下情况下,为什么回音后不插入换行符?
我有以下代码行:Php 在以下情况下,为什么回音后不插入换行符?,php,string,newline,line-breaks,string-concatenation,Php,String,Newline,Line Breaks,String Concatenation,我有以下代码行: $value['street1'] = "MCN"; $value['street2'] = "Bhavani peth spur"; $value['city'] = "Los Angeles"; $value['state_code'] = "CA"; $value['zip_code'] = 90009; $temp_rebate_data['user_address'] = $value['street1']."".($value['street2'] ? "\n
$value['street1'] = "MCN";
$value['street2'] = "Bhavani peth spur";
$value['city'] = "Los Angeles";
$value['state_code'] = "CA";
$value['zip_code'] = 90009;
$temp_rebate_data['user_address'] = $value['street1']."".($value['street2'] ? "\n".$value['street2'] : '')."\n".$value['city']."".$value['state_code']."-".$value['zip_code'];
echo $temp_rebate_data['user_address'];
die;
以上代码的输出如下:
MCN Bhavani peth spur Los AngelesCA-90009
预期产出应如下所示:
MCN,
Bhavani peth spur,
Los Angeles, CA - 90009
有人能帮我找出哪里做错了吗?这应该适合你:
$temp_rebate_data['user_address'] = $value['street1'].",".($value['street2'] ? "<br />".$value['street2'] . "," : '')."<br />".$value['city'].", ".$value['state_code']." - ".$value['zip_code'];
$temp\u retriet\u data['user\u address']=$value['street1'],“($value['street2']?”
“$value['street2'],”:”。“
“$value['city'],“$value['state\u code']。”-“$value['zip\u code'”;
只需在echo语句中的新行字符中使用shift+enter键即可使用
也可代替“使用”,或使用echo nl2br($data)代码>谢谢你的回答,但是前两行末尾和最后一行第一个字符串之后的逗号呢?