2015-10-15 91 views
-2

我有一段時間PHP statment應打印的喜歡的東西庫MySQLi只返回一個值

$full = "SELECT * FROM $table WHERE question LIKE '%$question%'"; 
$fullQ = mysqli_query($conn, $full); 
while($row = mysqli_fetch_assoc($fullQ)) { 
printf(nl2br("%s\n"), $row['question']); 
} 

但它打印結果只有一個使用UTF-8
我逃過了數據,但沒有

所有問題
+2

爲什麼它應該打印更多? –

+2

您的輸入是什麼?你是否檢查過數據庫實際上有多個匹配的行? – samlev

+0

是的,它有與同樣的$問題名稱的maaaaaany值 –

回答

-1

您是否嘗試在$ full之前放置一個真正的轉義字符串?

$question = mysqli->real_escape_string($question); 
$full = "SELECT * FROM $table WHERE question LIKE '%$question%'"; 
$fullQ = mysqli_query($conn, $full); 
while($row = mysqli_fetch_assoc($fullQ)) { 
    printf(nl2br("%s\n"), $row['question']); 
} 
+0

它沒有工作.. –

+0

它仍然不起作用 –

+0

閱讀關於'fetch_row()'。 –