php - Mysql_fetch_assoc(): supplied argument is not a valid MySQL result … -


संभव डुप्लिकेट:

मैं जो कुछ सोच सकता था, मैंने कोशिश की, लेकिन मैं इस त्रुटि को जारी रखता हूं।

Mysql_fetch_assoc (): प्रदान की गई तर्क एक मान्य MySQL परिणाम संसाधन / url / पर पंक्ति 41 पर नहीं है 

  यदि ($ _POST [सबमिट करें] == "सबमिट करें") {$ sql = "तालिका में शामिल करें (` सामग्री`, `उपयोगकर्ता आईडी`,` टीटीएमई`) मूल्य ('$ _POST) [सामग्री] ',' '। $ User_id।' ',' ".time ()।" ') "; $ Res = mysql_query ($ sql, $ लिंक) या मर (mysql_error ()); / * (लाइन 41 निम्नलिखित है) * / जबकि ($ result = mysql_fetch_assoc ($ res)) {}}

मैंने त्रुटि को छूने की कोशिश की (कोई त्रुटि केवल चेतावनी छापती है ), मैंने क्वेरी को बदलने की कोशिश की, जो कुछ मैं सोच सकता था यह कोड ठीक काम करता है - यह क्लिक पर सम्मिलित करता है, सब कुछ ठीक है, सिर्फ यह चेतावनी दिख रही है। _। '

कोई भी विचार?

mysql_query ()

रिटर्न वैल्यू

सेलेक्ट के लिए, "पोस्ट-टेक्स्ट" आइटमप्रॉप = "टेक्स्ट">

अन्य प्रकार के एसक्यूएल स्टेटमेंट्स, INSERT, अद्यतन, डेलेटेड, ड्रॉप, आदि के लिए, IISL, आदि, mysql_query () त्रुटि पर सफलता या गलत पर TRUE देता है।

लौटा परिणाम संसाधन mysql_fetch_array (), और परिणाम तालिकाओं से निपटने के लिए, डेटा।

DELETE, INSERT, REPLACE, या UPDATE कथन से कितनी पंक्तियों को प्रभावित किया गया है यह जानने के लिए SELECT कथन या mysql_affected_rows () के लिए कितनी पंक्तियां लौटाई गईं, यह जानने के लिए mysql_num_rows () का उपयोग करें। / P> < P> mysql_query () भी असफल हो जायेगी और लौटाएगा यदि उपयोगकर्ता को क्वेरी द्वारा संदर्भित तालिका (ए) तक पहुंचने की अनुमति नहीं है।


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