致命错误：无法在％$名称％

$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);

引用传递参数2有一个人可以帮

来源

2013-07-17 Daniel Hedelund Jensen

是参数整数或字符串？ –

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

它的字符串的副本，但它不会工作，即使我只与1做？并且只有1个“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.'%');

来源

2013-07-17 08:18:13

当我这样做时，我得到这个致命错误：调用一个非对象的成员函数bind_param（） –

$ Statement-> bind_param（“ss”，'％'。$ sogning '％'， '％' $ sogning '％'）。; 给出与相同的结果$ Statement-> bind_param（“ss”，$ sogning，$ sogning）; 给了我 –

我的朋友帮我解决..这是答案 –

致命错误：无法在％$名称％

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