Php Echo返回数组而不是值
我的echo返回数组而不是值Php Echo返回数组而不是值,php,mysql,Php,Mysql,我的echo返回数组而不是值 $postscount = mysql_query('select count(authorid) from topics where authorid="'.$author.'" '); $posts = mysql_fetch_row($postscount); 这里是echo: { echo "Posts: ".$posts." "; } 我试过这个: { echo '<pre>'; print_r($posts); e
$postscount = mysql_query('select count(authorid) from topics where authorid="'.$author.'" ');
$posts = mysql_fetch_row($postscount);
这里是echo:
{
echo "Posts: ".$posts." ";
}
我试过这个:
{
echo '<pre>';
print_r($posts);
echo '</pre>';
}
使用print\r可重新运行以下命令:
排列
[0]=>73
还有var_dump
数组1{[0]=>string2 73}
试一试
echo $posts[0];
或者就像
echo '<pre>'.var_export($posts, true).'</pre>';
使用此选项打印整个阵列:
$postscount = mysql_query('select count(authorid) from topics where authorid="'.$author.'" ') or die(mysql_error());
$posts = mysql_fetch_row($postscount);
var_dump($posts);
echo "There are ".$posts[0]. " authors";
试试这个代码
$posts = $name = $mysqli->query('select count(authorid) as postcount from topics where authorid="'.$author.'" ')->fetch_object()->postcount;
echo $posts;
这里
将告诉您查询是否成功。使用mysqli
查看PHP.net上的更多信息使用print\r进行了尝试,得到了什么?也许您有一个空数组,尝试使用var\u dumpMy echo returns数组,这是正确的。您可以。它们不再得到维护,而是在使用。而是学习,并使用PDO或MySQLi.array 0=>73,是吗?预期结果,因此??我希望它返回73警告:C:\xampp\htdocs\Forum\read_topic.php中的非法字符串偏移量“countauthorid”,位于第100行7,然后删除“countauthorid”数组1{[0]=>string2 73}
or die(mysql_error())
$posts = $name = $mysqli->query('select count(authorid) as postcount from topics where authorid="'.$author.'" ')->fetch_object()->postcount;
echo $posts;