Question

我正在编写一个PHP应用程序，其中一个功能是能够查询所有表(gene_r, genes_newL_dn, genes_newL_up, genes_oldL_up, genes_oldL_dn)中的基因交互。所以我有以下PHP函数来一次查询所有这些数据库。

public function getAllInteractions($input){
    $data = array();

    $sql_list = (
        "SELECT * FROM genes_r JOIN drugs_r ON drugs_r.id = genes_r.id WHERE drug_name='$input' OR gene_name='$input'",
        "SELECT * FROM genes_newL_dn JOIN drugs_newL_dn ON drugs_newL_dn.id = genes_newL_dn WHERE drug_name='$input' OR gene_name='$input'",
        "SELECT * FROM genes_newL_up JOIN drugs_newL_up ON drugs_newL_up.id = genes_newL_up WHERE drug_name='$input' OR gene_name='$input'",
        "SELECT * FROM genes_oldL_dn JOIN drugs_oldL_dn ON drugs_oldL_dn.id = genes_oldL_dn WHERE drug_name='$input' OR gene_name='$input'",
        "SELECT * FROM genes_oldL_up JOIN drugs_oldL_up ON drugs_oldL_up.id = genes_oldL_up WHERE drug_name='$input' OR gene_name='$input'"
    );

    foreach($sql_list as $sql){
        $query = $this->db->query($sql);
        // case 1 : SQL Query invalid / empty results
        if(!$query || $query->num_rows() == 0) {
            continue;   
        }
        else {
            $id = $query->row()->id;
            $sql = "SELECT interaction from matrix_r WHERE id='$id'";
            $query = $this->db->query($sql);
            array_push($data, $query->row());
        }   
    }

    return $data;
}

但是，我收到以下错误：

PHP Parse error:  syntax error, unexpected ',' in 
/chemicalgenomics/ci/application/models/search_model.php on line 81

我不明白为什么不喜欢'，'。它应该是一个字符串数组。有人可以帮帮我吗？感谢

PS：Line 81是数组中的第一行："SELECT * FROM ...

Answer 1

您初始化数组错误，您需要将array放在开头：

$sql_list = array(
    "SELECT * FROM genes_r JOIN drugs_r ON drugs_r.id = genes_r.id WHERE drug_name='$input' OR gene_name='$input'",
    "SELECT * FROM genes_newL_dn JOIN drugs_newL_dn ON drugs_newL_dn.id = genes_newL_dn WHERE drug_name='$input' OR gene_name='$input'",
    "SELECT * FROM genes_newL_up JOIN drugs_newL_up ON drugs_newL_up.id = genes_newL_up WHERE drug_name='$input' OR gene_name='$input'",
    "SELECT * FROM genes_oldL_dn JOIN drugs_oldL_dn ON drugs_oldL_dn.id = genes_oldL_dn WHERE drug_name='$input' OR gene_name='$input'",
    "SELECT * FROM genes_oldL_up JOIN drugs_oldL_up ON drugs_oldL_up.id = genes_oldL_up WHERE drug_name='$input' OR gene_name='$input'"
);

如果您使用的是PHP 5.4或更高版本（但并不是那么多的Web主机），您可以使用您接近的新的短数组语法，但该字符是括号（[）而不是括号（(）：

$sql_list = [
    "SELECT ...",
    "SELECT ...",
];

Answer 2

$sql_list = array(x, y, z);是PHP数组的正确语法。因此，在您的代码中，它应该是：

$sql_list = array(
    "SELECT * FROM genes_r JOIN drugs_r ON drugs_r.id = genes_r.id WHERE drug_name='$input' OR gene_name='$input'",
    "SELECT * FROM genes_newL_dn JOIN drugs_newL_dn ON drugs_newL_dn.id = genes_newL_dn WHERE drug_name='$input' OR gene_name='$input'",
    "SELECT * FROM genes_newL_up JOIN drugs_newL_up ON drugs_newL_up.id = genes_newL_up WHERE drug_name='$input' OR gene_name='$input'",
    "SELECT * FROM genes_oldL_dn JOIN drugs_oldL_dn ON drugs_oldL_dn.id = genes_oldL_dn WHERE drug_name='$input' OR gene_name='$input'",
    "SELECT * FROM genes_oldL_up JOIN drugs_oldL_up ON drugs_oldL_up.id = genes_oldL_up WHERE drug_name='$input' OR gene_name='$input'"
);

如果您使用的是PHP 5.4，则可以使用缩短的语法：

$sql_list = [
    "SELECT * FROM genes_r JOIN drugs_r ON drugs_r.id = genes_r.id WHERE drug_name='$input' OR gene_name='$input'",
    "SELECT * FROM genes_newL_dn JOIN drugs_newL_dn ON drugs_newL_dn.id = genes_newL_dn WHERE drug_name='$input' OR gene_name='$input'",
    "SELECT * FROM genes_newL_up JOIN drugs_newL_up ON drugs_newL_up.id = genes_newL_up WHERE drug_name='$input' OR gene_name='$input'",
    "SELECT * FROM genes_oldL_dn JOIN drugs_oldL_dn ON drugs_oldL_dn.id = genes_oldL_dn WHERE drug_name='$input' OR gene_name='$input'",
    "SELECT * FROM genes_oldL_up JOIN drugs_oldL_up ON drugs_oldL_up.id = genes_oldL_up WHERE drug_name='$input' OR gene_name='$input'"
];

创建数组php的问题

2 个答案: