致命錯誤：無法在％$名稱％

$query = " 
     SELECT 
      id, 
      overskrift, 
      tekst, 
      detaljer, 
      varenummer, 
      lager, 
      vegt, 
      pris, 
      billede, 
      fallow_id 
     FROM davidsen_vare 
     WHERE 
      overskrift LIKE ? OR varenummer LIKE ?) 
     LIMIT 30"; 

    $Statement = $this->mysqli->prepare($query); 
    $Statement->bind_param("ss","%$sogning%","%$sogning%"); //error here 
    $Statement->execute(); 
    $Statement->bind_result($id,$overskrift,$tekst,$detaljer,$varenummer,$lager,$vegt,$pris,$billede,$fallow_id);

引用傳遞參數2有一個人可以幫

來源

2013-07-17 Daniel Hedelund Jensen

是參數整數或字符串？ –

看起來像http://stackoverflow.com/questions/13105373/php-error-cannot-pass-parameter-2-by-reference –

它的字符串的副本，但它不會工作，即使我只與1做？並且只有1個「s」，％$ sogning％ –

$query = " 
SELECT 

    id, 
    overskrift, 
    tekst, 
    detaljer, 
    varenummer, 
    lager, 
    vegt, 
    pris, 
    billede, 
    fallow_id 
FROM davidsen_vare 

WHERE 
(overskrift LIKE %?% 
    OR varenummer LIKE %?%) 
    LIMIT 30"; 

    $Statement = $this->mysqli->prepare($query); 
    $Statement->bind_param("ss",$sogning,$sogning);

$query = " 
SELECT 

    id, 
    overskrift, 
    tekst, 
    detaljer, 
    varenummer, 
    lager, 
    vegt, 
    pris, 
    billede, 
    fallow_id 
FROM davidsen_vare 

WHERE 
(overskrift LIKE ? 
    OR varenummer LIKE ?) 
    LIMIT 30"; 

    $Statement = $this->mysqli->prepare($query); 
    $Statement->bind_param("ss",'%'.$sogning.'%','%'.$sogning.'%');

來源

2013-07-17 08:18:13

當我這樣做時，我得到這個致命錯誤：調用一個非對象的成員函數bind_param（） –

$ Statement-> bind_param（「ss」，'％'。$ sogning '％'， '％' $ sogning '％'）。; 給出與相同的結果$ Statement-> bind_param（「ss」，$ sogning，$ sogning）; 給了我 –

我的朋友幫我解決..這是答案 –

致命錯誤：無法在％$名稱％

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