Question

我试图将一部分文件名作为参数传递给我的php中的sql查询。我的文件名在表单中 db_feature_T1..iOXXdrXXcAMEra_092521.txt 从中我想要除db_feature和.txt之外的部分，即 T1..iOXXdrXXcAMEra_092521

我想要数据库中的img_id，其img_path = dressimages / T1..iOXXdrXXcAMEra_092521.jpg

我的代码如下。我能够分隔文件名，但我无法从数据库中获得任何结果。

            <?php
    include('db.php');

    if ($handle = opendir('C:\Users\adithi.a\Desktop\db_features')) {



        while (false !== ($entry = readdir($handle))) {
            $entry=substr($entry,11,-4); //this seperates the file name db_feature_T1..iOXXdrXXcAMEra_092521.txt to T1..iOXXdrXXcAMEra_092521



            $image_path='dressimages/'.$entry.'.jpg';//I want to pass the img_path as it is saved in the database in the form of dressimages/T1..iOXXdrXXcAMEra_092521.jpg


         $result=  mysql_query("select img_id from tbl_image where img_path='$image_path'") or die("Sorry");
          echo $result;//I dont get anything as output. 
        }

    }

        closedir($handle);
    ?>

我在执行上面的代码时进入了无限循环，所以我尝试了：

    $image_path='dressimages/'.$entry.'.jpg';//I want to pass the img_path as it is saved in the database in the form of dressimages/T1..iOXXdrXXcAMEra_092521.jpg

        $sql = "select img_id from tbl_image where img_path=".$image_path;
     echo $sql . '<br />';
        $result=mysql_query("$sql");

       }

            while($data=mysql_fetch_assoc($result))
                  {

                   echo $data["img_id"];
                   }

现在我得到错误mysql_fetch_assoc（）期望参数1是资源，布尔值在第34行的C：\ wamp \ www \ FashionBonanza \ Readfile.php中给出

任何帮助

Answer 1

您必须先fetch结果中的数据

$data=mysql_fetch_assoc($result);  // this is missing
print_r($data); // or 
echo $data["img_id"];

如果可以有多个结果，你可以循环

while($data=mysql_fetch_assoc($result))
{
print_r($data); // or 
echo $data["img_id"];
}

Answer 2

因为你正在处理这么多文件，我会尝试另一种方法。那就是将所有路径存储到一个数组中，进行单个查询并将id与路径相关联。我不确定这段代码是否有效（我还没有测试过），但我希望你能得到这样的代码：

<?php
include('db.php');

if ($handle = opendir('C:\Users\adithi.a\Desktop\db_features')) {

//Store all paths into $image_paths array
$image_paths = array();
while (false !== ($entry = readdir($handle))) {
    $entry=substr($entry,11,-4); //this seperates the file name db_feature_T1..iOXXdrXXcAMEra_092521.txt to T1..iOXXdrXXcAMEra_092521
    if (strlen($entry) !== 0) {
    $image_paths[]='dressimages/'.$entry.'.jpg';//I want to pass the img_path as it is saved in the database in the form of dressimages/T1..iOXXdrXXcAMEra_092521}
    }
}
closedir($handle);

//Implode paths
$pathQuotesArray = array_map('apply_quotes', $image_paths); //Looks like 'filename1', 'filename2' etc
$pathQuotes = implode(',', $pathQuotesArray); 

//Do one query 
$result=  mysql_query("select img_id from tbl_image WHERE img_path IN ($pathQuotes)") or die(mysql_error());

//Associate id's with paths
while($data=mysql_fetch_assoc($result))
{
    $key[$data["img_id"]] = $data["img_path"];
}

echo $key[5]; //If img_id is 5, then it would show correct path (hopefully :-))

function apply_quotes($item)
{
    return "'" . mysql_real_escape_string($item) . "'";
}



?>

Answer 3

试试这个

$result=  mysql_query("select img_id from tbl_image where img_path="$image_path) or die("Sorry");

Answer 4

只有两个想法：

$result=  mysql_query("select img_id from tbl_image where img_path='".$image_path."'") or die("Sorry");

也许以这种方式做得更好。否则为SQL查询设置一个额外的字符串变量，这样就可以跟踪已执行的确切查询。

$sQry = "select img_id from tbl_image where img_path='".$image_path."'";
var_dump($sQry);
$result=  mysql_query($sQry) or die("Sorry");

因此，您可以检查变量是否包含正确的值。

下一点是“回声”。您无法回显结果集，使用mysql-fetch-array（）或mysql_fetch_assoc（） - http://php.net/manual/de/function.mysql-fetch-array.php等函数或其他函数。

while ($row = mysql_fetch_assoc($result)) {
    echo $row["img_id "];
}

将php变量传递给mysql查询时无法获得结果

4 个答案: