从表中选择*，其中名称为％O％

Question

In phpMyAdmin the query

SELECT * FROM table WHERE value1 LIKE '%O%' OR value2 LIKE '%1%'

nicely returns all rows containing O or 1.

But the same query implemented in a PHP script only returns the first row matching O or 1. Instead off all rows matching O or 1.

This is my current PHP script:

mysqli_select_db ($con, $database) or die (mysqli_error());

            $sqli = mysqli_query ($con,"SELECT * FROM table WHERE VALUE1 LIKE '%O%' OR VALUE2 LIKE '%1%'") or die (mysqli_error()); 

            $echo = mysqli_fetch_assoc($sqli);

            if (empty ($echo))
            {
                echo "No matching results found.";
            }
            else
            {
                echo "<table width='50%' border='1'>";
                echo "<tr><th>VALUE1</th><th>VALUE2</th></tr>";
                echo "<tr><td>{$echo["VALUE1"]}</td>";
                echo "<td>{$echo["VALUE2"]}</td></tr>";
            }

The output is only the first matching row. Instead of all matching rows. Also when I use print_r instead of table rows.

Answer 1

You have to iterate through all the rows with a loop like this

if( !mysqli_num_rows($sqli) )
{
    echo "No matching results found.";
} else {
    $table = "<table width='50%' border='1'>
";
    $table .= "<tr><th>VALUE1</th><th>VALUE2</th></tr>
";
    while( $echo = mysqli_fetch_assoc($sqli) )
    {
        $table .= "<tr><td>{$echo['VALUE1']}</td><td>{$echo['VALUE2']}</td></tr>
";
    }
    $table .= "</table>";
}

Note I added a couple of lines of code to count the rows and display a message if there are no matches found (as I intended is your goal from your question).

---

Edit: Better would be to use the built in num_rows instead of reinventing the wheel

---

Edit 2: I have a feeling he didn't want the table to duplicate for each row. Change the code a bit