dsomm80482 2015-11-13 13:54
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MySQL不会检索PHP中的第一项

I've now been trying for hour and can't figure the problem out. I've made a php file that fetch all items in a table and retrieves that as JSON. But for some reason after I inserted the second mysql-query, it stopped fetching the first item. My code is following:

...
    case "LoadEntryList":
        $result2 = performquery("SELECT * FROM Entries WHERE Category = '" . $_POST["Category"] .
        "' LIMIT " . $_POST["Offset"] . ", " . $_POST["Quantity"] . "");
        $row2 = $result2->fetch_assoc();
    while($row = $result2->fetch_assoc()) {
        $result3 = performquery("SELECT Username FROM Users WHERE ID = '" . $row2["UserID"] . "'");
        $row3 = $result3->fetch_assoc();
   echo substr(json_encode($row),0,
        strlen(json_encode($row))-1) . ",\"Username\":\"" . $row3["Username"]  . "\"}";
}
...

Any help is greatly appreciated.

EDIT: Thanks for all those super fast responses.

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4条回答 默认 最新

  • duandou1903 2015-11-13 14:03
    关注

    No Need of $row2 = $result2->fetch_assoc();

        <?
        case "LoadEntryList":
        $result2 = performquery("SELECT * FROM Entries WHERE Category = '" . $_POST["Category"] .
            "' LIMIT " . $_POST["Offset"] . ", " . $_POST["Quantity"] . "");
        while($row = $result2->fetch_assoc()) 
        {
            $result3 = performquery("SELECT Username FROM Users WHERE ID = '" . $row["UserID"] . "'");
            $row3 = $result3->fetch_assoc();
            echo substr(json_encode($row),0,strlen(json_encode($row))-1) . ",\"Username\":\"" . $row3["Username"]  . "\"}";
        }
        ?>
    

    Or,

    <?
    ...
        case "LoadEntryList":
        $Category=$_POST["Category"];
        $Offset=$_POST["Offset"];
        $Quantity=$_POST["Quantity"];
    
        $result3 = performquery("SELECT Entries.*, Users.Username FROM Entries, Users WHERE Entries.Category=$Category AND Entries.UserID=Users.ID LIMIT $Offset, $Quantity");
    
        $row3 = $result3->fetch_assoc();
        echo substr(json_encode($row),0,strlen(json_encode($row))-1) . ",\"Username\":\"" . $row3["Username"]  . "\"}";
    }
    ...
    ?>
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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