Question

function get_subject_by_id($subject_id){
    global $connection; 
    $query = "SELECT * ";
    $query .= "FROM subjects ";
    $query .= "WHERE 'id' ={$subject_id} "; 
    $query .= "LIMIT 1 "; 
    $result_set = mysql_query($query, $connection); 
    confirm_query($result_set); 
    // if no rows are returned fetch_array will return false
    if(!is_null($result_set)){
    echo "got here";
    echo $result_set; 
    } 
    if($subject=mysql_fetch_array($result_set)){ // not working properly
        echo "testing fetch array";
        return $subject; 
    }else {
        return NULL; 
    }

虽然我的表中有3个项目，但fetch查询仍会返回false。一些代码显然正在测试是否我得到了结果并且这些都工作但是fetch数组没有执行if语句，因为它认为它是假的。请帮助几个小时的挫折属于这个问题

Answer 1

您的查询中的引号类型错误，它们应该是反引号：

$query .= "WHERE `id` ={$subject_id} "; 
                 ^  ^

引号使id成为字符串，而不是列名。由于字符串id永远不等于$subject_id，因此查询不会返回任何内容。

Answer 2

我认为这是因为id列用单引号括起来而不是后面的刻度

$query .= "WHERE 'id' ={$subject_id} ";

应该是

$query .= "WHERE id ={$subject_id} ";

或者

$query .= "WHERE `id` ={$subject_id} ";

mysql fetch数组继续返回false

2 个答案: