PHP mysql返回null，硬编码的SQL查询有效

Question

I'm having an issue in which I'm unable to get my database query within a PHP program to work. The query works fine within the program if it is hard coded, but it fails otherwise and passes back no results. I have echo'd the two results and am given a different string length, but the string I am given is identical (strings gotten via var_dump). I'm at my wit's end; I'm not really sure what is the issue with the query. I have tried several different fixes which I found for similar problems, but none of them have worked. I trim the posted input and also have the variable double quoted as opposed to single quoted so that the reference executes. I really just have no clue what's wrong. Here is the code that's relevant to this project: AJAX call to php class:

chlorinator = ($('#chlorinator').val()).concat(' GS').trim();
                $.ajax(
                {
            type: "POST",
            url: "gravity.php",
            data: "chlorinator="+chlorinator,
            cache: false,
            beforeSend: function () { 
            $('#results').html('<img src="loader.gif" alt="" width="24" height="24">');
            },
            success: function(html) 
            {    
                $("#results").html( html );
            }});

And here is the relevant php code:

<?php
include 'connection.php';
$chlorinator = trim( mysqli_real_escape_string ($dbhandle,$_POST["chlorinator"]));


$query = 'SELECT chlorinators.model_name, equipment.name, equipment.cutsheet_url, chlorinators.pump_specific, equipment.file_name
            FROM chlorinators 
            INNER JOIN chlorinator_equipment
            ON chlorinators.chlorinator_index = chlorinator_equipment.chlorinator_index
            INNER JOIN equipment
            ON chlorinator_equipment.equipment_index = equipment.equipment_index    
            WHERE chlorinators.model_name= "' . $chlorinator . '"';

echo "The value of the combined string is:<br> ";               
var_dump($query);
echo '<br><br>';

echo "The value of the hard-coded string is:<br> ";
$query = 'SELECT chlorinators.model_name, equipment.name, equipment.cutsheet_url, chlorinators.pump_specific, equipment.file_name
            FROM chlorinators 
            INNER JOIN chlorinator_equipment
            ON chlorinators.chlorinator_index = chlorinator_equipment.chlorinator_index
            INNER JOIN equipment
            ON chlorinator_equipment.equipment_index = equipment.equipment_index    
            WHERE chlorinators.model_name= "2075 GS"';
            var_dump($query);
echo '<br><br>';

if ($result = $dbhandle->query($query))
{?>
<br><br><?php
var_dump($result->fetch_assoc());

printf("<p style='font-family:sans-serif; text-align: center;'>The components of the %s are listed below</p><table id='form' name='pump' style='margin: auto; padding: auto'>", $_POST["chlorinator"]);

while($row = $result->fetch_assoc())
{ 
    printf ("<div><tr><td>%s</td><td><a href='%s' download>Download</a></td></tr>", $row["name"],$row["cutsheet_url"]); 
}
printf('</table>');
}


?>

For this particular example I'm using the value '2075 GS' as the chlorinator value. It is generally passed via change on a selection box, so the values are hard coded and correct. The output of this specific example is:

string(403) "SELECT chlorinators.model_name, equipment.name, equipment.cutsheet_url, chlorinators.pump_specific, equipment.file_name FROM chlorinators INNER JOIN chlorinator_equipment ON chlorinators.chlorinator_index = chlorinator_equipment.chlorinator_index INNER JOIN equipment ON chlorinator_equipment.equipment_index = equipment.equipment_index WHERE chlorinators.model_name= "2075 GS""

string(404) "SELECT chlorinators.model_name, equipment.name, equipment.cutsheet_url, chlorinators.pump_specific, equipment.file_name FROM chlorinators INNER JOIN chlorinator_equipment ON chlorinators.chlorinator_index = chlorinator_equipment.chlorinator_index INNER JOIN equipment ON chlorinator_equipment.equipment_index = equipment.equipment_index WHERE chlorinators.model_name= "2075 GS""

I don't see any difference between the two outputs; any idea as to where the one character difference is and how I can eliminate it so that my query will properly work? Any help is greatly appreciated.

Answer 1

Use double quoted string to build these query striings its so much easier as $variables are then automatically expanded and you dont have to fiddle with . concatenation at all

$query = "SELECT chlorinators.model_name, equipment.name,
                 equipment.cutsheet_url, chlorinators.pump_specific, 
                 equipment.file_name
          FROM chlorinators 
             INNER JOIN chlorinator_equipment 
                ON chlorinators.chlorinator_index = chlorinator_equipment.chlorinator_index
             INNER JOIN equipment
                ON chlorinator_equipment.equipment_index = equipment.equipment_index    
        WHERE chlorinators.model_name= '$chlorinator'";

Of course it could be an annoying byte corruption in your source file. I cant count the hours I have wasted trying to work out why code wont compile when it was just a byte corruption in the source. In that case copy and paste the code to a new file, then delete the old file and rename your new file to the old filename

I think the reason you are seeing no data in the page is that you have read it and thrown it away

if ($result = $dbhandle->query($query)) {
    echo '<br><br>';

    // this line read your result row and ignored it
    //var_dump($result->fetch_assoc());

    printf("<p style='font-family:sans-serif; text-align: center;'>The components of the %s are listed below</p><table id='form' name='pump' style='margin: auto; padding: auto'>", $_POST["chlorinator"]);

    // now the data should be available to read here
    // I assume you only have one row with 
    // chlorinators.model_name= "2075 GS"

    while($row = $result->fetch_assoc()) { 
        printf ("<div><tr><td>%s</td><td><a href='%s' download>Download</a></td></tr>", $row["name"],$row["cutsheet_url"]); 
    }
    printf('</table>');
} else {

    // there must have been an erro in the query we just executed
    // so show it.
    echo $dbhandle->error;
    exit;

?>

Answer 2

You should try debugging with the proper functions like this:

echo 'MySQL reports error #'.mysqli_errno($dbhandle).' - '.mysqli_error($dbhandle);

ps Looking at your code, I cannot spot your error.