mysql_fetch_array返回一个数组，数字为“1”

I have a problem with mysql_fetch_array. I'm trying to do this:

我有mysql_fetch_array的问题。我正在尝试这样做：

if(count(mysql_fetch_array(mysql_query($peticion)))) {
    array_push($errores, $texto['conta_existe']);
}

The previous code should detect if mysql_fetch_array found a row in my database and, if it found at least one, an array should be pushed. Problem is that if I haven't got anything in my db, fetch_array returns the number "1".

前面的代码应检测mysql_fetch_array是否在我的数据库中找到了一行，如果找到至少一行，则应该推送一个数组。问题是，如果我的数据库中没有任何内容，则fetch_array返回数字“1”。

I tried to find what's happening with this peace of code:

我试图通过这种代码安静来发现正在发生的事情：

$arrayc = mysql_fetch_array(mysql_query($peticion), MYSQL_NUM);
echo 'PRINT_R: ';
echo print_r($arrayc);
echo '<br>COUNT: ';
echo count($arrayc);

And returns this:

并返回：

PRINT_R: Array ( [0] => FLEREX [conta] => FLEREX ) 1

PRINT_R：数组（[0] => FLEREX [conta] => FLEREX）1

COUNT: 2

COUNT：2

I don't understand why there's that number one there, after the array. The previous quote was returned with only one row, to show you the array; but if there's not any row in the db, this is what I get:

我不明白为什么在阵列之后那里有第一号。之前的引用仅返回一行，以显示数组;但如果数据库中没有任何行，这就是我得到的：

PRINT_R: 1

PRINT_R：1

COUNT: 1

COUNT：1

I don't know where comes from that one, but is always there.

我不知道那个来自哪里，但总是在那里。

Thanks for reading and sorry for my bad english.

感谢您阅读并抱歉我的英语不好。

2 个解决方案

#1

Try mysql_num_rows instead of count and delete the mysql_fetch_array

尝试mysql_num_rows而不是count并删除mysql_fetch_array

2 个解决方案

#1

更多相关文章

随机推荐