Question

我有以下搜索代码段，当我仅将其与mysql表中的value1和value2数据一起使用时，它可以工作。

当我将Mysql CASE Function添加到具有value1r和value2r的查询中时（如以下示例所示），并尝试在页面上进行搜索时，它给了我这个错误"Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in...."，这意味着我有一个错误在查询中。

<?php

if(isset($_POST['search']))
{
    $valueToSearch = $_POST['valueToSearch'];

    $query = "SELECT * FROM gdata 
    WHERE            
    Value1 = '" . $valueToSearch . "' OR
    Value2 = '" . $valueToSearch . "' OR
    Value1r = '" . $valueToSearch . "' OR
    Value2r = '" . $valueToSearch . "' 
    ";

    $search_result = filterTable($query);

}
 else {
    $query = "SELECT Value1, Value2,

  CASE
                 WHEN Value1 >= 90 AND Value1 <= 100
                 THEN 'A'
                 WHEN Value1 >= 80 AND Value1 <= 89
                 THEN 'B'
                 WHEN Value1 >= 70 AND Value1 <= 79
                 THEN 'C'  
                 WHEN Value1 >= 0 AND Value1 <= 69
                 THEN 'F'
                 ELSE '' END AS Value1r,

                   CASE
                 WHEN Value2 >= 90 AND Value2 <= 100
                 THEN 'A'
                 WHEN Value2 >= 80 AND Value2 <= 89
                 THEN 'B'
                 WHEN Value2 >= 70 AND Value2 <= 79
                 THEN 'C'  
                 WHEN Value2 >= 0 AND Value2 <= 69
                 THEN 'F'
                 ELSE '' END AS Value2r 

  FROM gdata";


    $search_result = filterTable($query);
}

// function to connect and execute the query
function filterTable($query)
{
    $connect = mysqli_connect("localhost", "root", "", "db");
    $filter_Result = mysqli_query($connect, $query);
    return $filter_Result;
}

?>

<!DOCTYPE html>
<html>
    <head>
        <title>Gdata</title>
</head>

    <body>


<form action="gdata.php" method="post">
<input type="text" name="valueToSearch" placeholder="Search"> 
<input type="submit" name="search" value="Filter"> 

<table> 
<tr>
   <th>G1</th> 
   <th>G2</th>
   <th>R1</th>
   <th>R2</th>
</tr>

<?php 
    while($row = mysqli_fetch_array($search_result))
      { 
  echo "<tr>"; 
  echo "<td>" . $row['Value1'] . "</td>"; 
  echo "<td>" . $row['Value2'] . "</td>"; 
  echo "<td>" . $row['Value1r'] . "</td>";  
  echo "<td>" . $row['Value2r'] . "</td>";   
  echo "</tr>"; 
       } 

echo "</table>";
?>     
</form>

</body>
</html>

我在这里做错了什么？请给我一点帮助吗？谢谢！

注意：我的问题不是专门针对给定的错误。

Answer 1

此查询在我们的数据库上触发。

CREATE OR REPLACE VIEW gdata_view AS
SELECT Value1, Value2, CASE WHEN Value1 >= 90 AND Value1 <= 100 THEN 'A' WHEN Value1 >= 80 AND Value1 <= 89 THEN 'B' WHEN Value1 >= 70 AND Value1 <= 79 THEN 'C' WHEN Value1 >= 0 AND Value1 <= 69 THEN 'F' ELSE '' END AS Value1r, CASE WHEN Value2 >= 90 AND Value2 <= 100 THEN 'A' WHEN Value2 >= 80 AND Value2 <= 89 THEN 'B' WHEN Value2 >= 70 AND Value2 <= 79 THEN 'C' WHEN Value2 >= 0 AND Value2 <= 69 THEN 'F' ELSE '' END AS Value2r FROM gdata

create view for gdata table.
view is your temporary table.
when you change any data in main table view automatically affected.

when you submit data change your table name in your code gdata to gdata_view (gdata_view is a temporary table <= without view is not possible)

$query = "SELECT * FROM gdata 
    WHERE            
    Value1 = '" . $valueToSearch . "' OR
    Value2 = '" . $valueToSearch . "' OR
    Value1r = '" . $valueToSearch . "' OR
    Value2r = '" . $valueToSearch . "' 
    ";
replace by
$query = "SELECT * FROM gdata_view 
    WHERE            
    Value1 = '" . $valueToSearch . "' OR
    Value2 = '" . $valueToSearch . "' OR
    Value1r = '" . $valueToSearch . "' OR
    Value2r = '" . $valueToSearch . "' 
    ";

Answer 2

您可以使用以下内容来跟踪错误，请将此代码段放在mysqli_query

之后

if (!$filter_Result) {
    printf("Error: %s\n", mysqli_error($connect));
    exit();
}

Answer 3

此错误表示查询给出没有记录的空结果尝试再次编写查询，并在phpmyadmin中对其进行测试，如果它在那里工作正常，那么它将作为您的应用程序运行

PHP搜索无法与CASE函数一起使用

3 个答案:

此查询在我们的数据库上触发。